What, if anything, is the sum of all complex numbers?












13














If this question is ill-defined or is otherwise of poor quality, then I'm sorry.




What, if anything, is the sum of all complex numbers?




If anything at all, it's an uncountable sum, that's for sure.



I'm guessing some version of the Riemann series theorem would mean that there is no such thing as the sum of complex numbers, although - and I hesitate to add this - I would imagine that



$$sum_{zinBbb C}z=0tag{$Sigma$}$$



is, in some sense, what one might call the "principal value" of the sum. For all $winBbb C$, we have $-winBbb C$ with $w+(-w)=0$, so, if we proceed naïvely, we could say that we are summing $0$ infinitely many times${}^dagger$, hence $(Sigma)$.



We have to make clear what we mean by "sum", though, since, of course, with real numbers, one can define all sorts of different types of infinite sums. I'm at a loss here.



Has this sort of thing been studied before?



I'd be surprised if not.



Please help :)





$dagger$ I'm well aware that this is a bit too naïve. It's not something I take seriously.










share|cite|improve this question




















  • 2




    Since each $x in mathbb{C}$ has its corresponding $-x$, the sum must vanish.
    – David G. Stork
    21 hours ago








  • 2




    There's no order to complex numbers, so is there any sense of it converging in the first place? I'm not sure, @DietrichBurde
    – Shaun
    21 hours ago






  • 9




    I am not aware of any definition of a sum of an uncountable set.
    – Lee Mosher
    21 hours ago






  • 8




    @DavidG.Stork This is no argument, see Grandi's series.
    – Dietrich Burde
    21 hours ago








  • 3




    @LeeMosher My point was that you'd said you were not aware of any definition. There is a fairly standard and natural definition of $sum_{zinBbb C} c_z$, according to which $sum_{zinBbb C} z$ does not exist - that seems more satisfactory then jjust saying we don't have a definition. (It really is the "right" definition imo, being exactly the limit of the net of partial sums, $lim_{FtoBbb C}sum_F$. Turns out to be equivalent to integrability with respect to counting measure...)
    – David C. Ullrich
    17 hours ago
















13














If this question is ill-defined or is otherwise of poor quality, then I'm sorry.




What, if anything, is the sum of all complex numbers?




If anything at all, it's an uncountable sum, that's for sure.



I'm guessing some version of the Riemann series theorem would mean that there is no such thing as the sum of complex numbers, although - and I hesitate to add this - I would imagine that



$$sum_{zinBbb C}z=0tag{$Sigma$}$$



is, in some sense, what one might call the "principal value" of the sum. For all $winBbb C$, we have $-winBbb C$ with $w+(-w)=0$, so, if we proceed naïvely, we could say that we are summing $0$ infinitely many times${}^dagger$, hence $(Sigma)$.



We have to make clear what we mean by "sum", though, since, of course, with real numbers, one can define all sorts of different types of infinite sums. I'm at a loss here.



Has this sort of thing been studied before?



I'd be surprised if not.



Please help :)





$dagger$ I'm well aware that this is a bit too naïve. It's not something I take seriously.










share|cite|improve this question




















  • 2




    Since each $x in mathbb{C}$ has its corresponding $-x$, the sum must vanish.
    – David G. Stork
    21 hours ago








  • 2




    There's no order to complex numbers, so is there any sense of it converging in the first place? I'm not sure, @DietrichBurde
    – Shaun
    21 hours ago






  • 9




    I am not aware of any definition of a sum of an uncountable set.
    – Lee Mosher
    21 hours ago






  • 8




    @DavidG.Stork This is no argument, see Grandi's series.
    – Dietrich Burde
    21 hours ago








  • 3




    @LeeMosher My point was that you'd said you were not aware of any definition. There is a fairly standard and natural definition of $sum_{zinBbb C} c_z$, according to which $sum_{zinBbb C} z$ does not exist - that seems more satisfactory then jjust saying we don't have a definition. (It really is the "right" definition imo, being exactly the limit of the net of partial sums, $lim_{FtoBbb C}sum_F$. Turns out to be equivalent to integrability with respect to counting measure...)
    – David C. Ullrich
    17 hours ago














13












13








13


6





If this question is ill-defined or is otherwise of poor quality, then I'm sorry.




What, if anything, is the sum of all complex numbers?




If anything at all, it's an uncountable sum, that's for sure.



I'm guessing some version of the Riemann series theorem would mean that there is no such thing as the sum of complex numbers, although - and I hesitate to add this - I would imagine that



$$sum_{zinBbb C}z=0tag{$Sigma$}$$



is, in some sense, what one might call the "principal value" of the sum. For all $winBbb C$, we have $-winBbb C$ with $w+(-w)=0$, so, if we proceed naïvely, we could say that we are summing $0$ infinitely many times${}^dagger$, hence $(Sigma)$.



We have to make clear what we mean by "sum", though, since, of course, with real numbers, one can define all sorts of different types of infinite sums. I'm at a loss here.



Has this sort of thing been studied before?



I'd be surprised if not.



Please help :)





$dagger$ I'm well aware that this is a bit too naïve. It's not something I take seriously.










share|cite|improve this question















If this question is ill-defined or is otherwise of poor quality, then I'm sorry.




What, if anything, is the sum of all complex numbers?




If anything at all, it's an uncountable sum, that's for sure.



I'm guessing some version of the Riemann series theorem would mean that there is no such thing as the sum of complex numbers, although - and I hesitate to add this - I would imagine that



$$sum_{zinBbb C}z=0tag{$Sigma$}$$



is, in some sense, what one might call the "principal value" of the sum. For all $winBbb C$, we have $-winBbb C$ with $w+(-w)=0$, so, if we proceed naïvely, we could say that we are summing $0$ infinitely many times${}^dagger$, hence $(Sigma)$.



We have to make clear what we mean by "sum", though, since, of course, with real numbers, one can define all sorts of different types of infinite sums. I'm at a loss here.



Has this sort of thing been studied before?



I'd be surprised if not.



Please help :)





$dagger$ I'm well aware that this is a bit too naïve. It's not something I take seriously.







sequences-and-series analysis complex-numbers summation definition






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 19 hours ago

























asked 21 hours ago









Shaun

8,452113580




8,452113580








  • 2




    Since each $x in mathbb{C}$ has its corresponding $-x$, the sum must vanish.
    – David G. Stork
    21 hours ago








  • 2




    There's no order to complex numbers, so is there any sense of it converging in the first place? I'm not sure, @DietrichBurde
    – Shaun
    21 hours ago






  • 9




    I am not aware of any definition of a sum of an uncountable set.
    – Lee Mosher
    21 hours ago






  • 8




    @DavidG.Stork This is no argument, see Grandi's series.
    – Dietrich Burde
    21 hours ago








  • 3




    @LeeMosher My point was that you'd said you were not aware of any definition. There is a fairly standard and natural definition of $sum_{zinBbb C} c_z$, according to which $sum_{zinBbb C} z$ does not exist - that seems more satisfactory then jjust saying we don't have a definition. (It really is the "right" definition imo, being exactly the limit of the net of partial sums, $lim_{FtoBbb C}sum_F$. Turns out to be equivalent to integrability with respect to counting measure...)
    – David C. Ullrich
    17 hours ago














  • 2




    Since each $x in mathbb{C}$ has its corresponding $-x$, the sum must vanish.
    – David G. Stork
    21 hours ago








  • 2




    There's no order to complex numbers, so is there any sense of it converging in the first place? I'm not sure, @DietrichBurde
    – Shaun
    21 hours ago






  • 9




    I am not aware of any definition of a sum of an uncountable set.
    – Lee Mosher
    21 hours ago






  • 8




    @DavidG.Stork This is no argument, see Grandi's series.
    – Dietrich Burde
    21 hours ago








  • 3




    @LeeMosher My point was that you'd said you were not aware of any definition. There is a fairly standard and natural definition of $sum_{zinBbb C} c_z$, according to which $sum_{zinBbb C} z$ does not exist - that seems more satisfactory then jjust saying we don't have a definition. (It really is the "right" definition imo, being exactly the limit of the net of partial sums, $lim_{FtoBbb C}sum_F$. Turns out to be equivalent to integrability with respect to counting measure...)
    – David C. Ullrich
    17 hours ago








2




2




Since each $x in mathbb{C}$ has its corresponding $-x$, the sum must vanish.
– David G. Stork
21 hours ago






Since each $x in mathbb{C}$ has its corresponding $-x$, the sum must vanish.
– David G. Stork
21 hours ago






2




2




There's no order to complex numbers, so is there any sense of it converging in the first place? I'm not sure, @DietrichBurde
– Shaun
21 hours ago




There's no order to complex numbers, so is there any sense of it converging in the first place? I'm not sure, @DietrichBurde
– Shaun
21 hours ago




9




9




I am not aware of any definition of a sum of an uncountable set.
– Lee Mosher
21 hours ago




I am not aware of any definition of a sum of an uncountable set.
– Lee Mosher
21 hours ago




8




8




@DavidG.Stork This is no argument, see Grandi's series.
– Dietrich Burde
21 hours ago






@DavidG.Stork This is no argument, see Grandi's series.
– Dietrich Burde
21 hours ago






3




3




@LeeMosher My point was that you'd said you were not aware of any definition. There is a fairly standard and natural definition of $sum_{zinBbb C} c_z$, according to which $sum_{zinBbb C} z$ does not exist - that seems more satisfactory then jjust saying we don't have a definition. (It really is the "right" definition imo, being exactly the limit of the net of partial sums, $lim_{FtoBbb C}sum_F$. Turns out to be equivalent to integrability with respect to counting measure...)
– David C. Ullrich
17 hours ago




@LeeMosher My point was that you'd said you were not aware of any definition. There is a fairly standard and natural definition of $sum_{zinBbb C} c_z$, according to which $sum_{zinBbb C} z$ does not exist - that seems more satisfactory then jjust saying we don't have a definition. (It really is the "right" definition imo, being exactly the limit of the net of partial sums, $lim_{FtoBbb C}sum_F$. Turns out to be equivalent to integrability with respect to counting measure...)
– David C. Ullrich
17 hours ago










3 Answers
3






active

oldest

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19














Traditionally, the sum of a sequence is defined as the limit of the partial sums; that is, for a sequence ${a_n}$, $sum{a_n}$ is that number $S$ so that for every $epsilon > 0$, there is an $N$ such that whenever $m > N$, $|S - sum_{n = 0}^ma_n| < epsilon$. There's no reason we can't define it like that for uncountable sequences as well: let $mathfrak{c}$ be the cardinality of $mathbb{C}$, and let ${a_{alpha}}$ be a sequence of complex numbers where the indices are ordinals less than $mathfrak{c}$. We define $sum{a_{alpha}}$ as that value $S$ so that for every $epsilon > 0$, there is a $beta < mathfrak{c}$ so that whenever $gamma > beta$, $|S - sum_{alpha = 0}^{gamma}a_{alpha}| < epsilon$. Note that this requires us to recursively define transfinite summation, to make sense of that sum up to $gamma$.



But here's the thing: taking $epsilon$ to be $1$, then $1/2$, then $1/4$, and so on, we get a sequence of "threshold" $beta$ corresponding to each one; call $beta_n$ the $beta$ corresponding to $epsilon = 1/2^n$. This is a countable sequence (length strictly less than $mathfrak{c}$). Inconveniently, $mathfrak{c}$ is regular: any increasing sequence of ordinals less than $mathfrak{c}$ with length less than $mathfrak{c}$ must be bounded strictly below $mathfrak{c}$. So that means there's some $beta_{infty}$ that's below $mathfrak{c}$ but greater than every $beta_n$. But by definition, that means that all partial sums past $beta_{infty}$ are less than $1/2^n$ away from $S$ for every $n$. So they must be exactly equal to $S$. And that means that we must be only adding $0$ from that point forward.



This is a well-known result that I can't recall a reference for: the only uncountable sequences that have convergent sums are those which consist of countably many nonzero terms followed by nothing but zeroes. In other words, there's no sensible way to sum over all of the complex numbers and get a convergence.






share|cite|improve this answer

















  • 2




    How does one mentally reconcile this information with the idea of an integral? You could view that as a continuous sum.
    – orlp
    18 hours ago










  • That's a very good question, @orlp; I think I'll withdraw the acceptance of this answer until I understand what it implies about integrals.
    – Shaun
    18 hours ago






  • 1




    The regularity of $mathfrak{c}$ is really a red herring here (and in fact, $mathfrak{c}$ may not be regular!). You could just as well pick an enumeration of $mathbb{C}$ whose length has countable cofinality, and then there might not be any point beyond which all the terms are $0$. What you can prove though (independent of the ordering used) is that only countably many terms can be nonzero. (Proof: if there are uncountably many nonzero terms, apply your argument to the first partial sum which contains uncountably many nonzero terms, which must have cofinality $omega_1$.)
    – Eric Wofsey
    14 hours ago












  • As for the regularity of $mathfrak{c}$, it is consistent with ZFC that $mathfrak{c}$ is singular (for instance, it could be $aleph_{omega_1}$). However, it follows from König's theorem that it has uncountable cofiinality so your argument still does work for it.
    – Eric Wofsey
    14 hours ago






  • 3




    @orlp An integral isn't an infinite sum; it's an infinite average. You're effectively dividing each term by a measure of the "number" of terms, rendering it infinitesimal; in this case, the uncountable analogue of infinitesimal. That means it doesn't really resemble a "sum" in the sense we usually mean (for example, the integral of the function $f(x)$ which is $1$ at $0$ and $1$ and zero elsewhere is not $2$).
    – Reese
    9 hours ago



















10














If ${z_i:iin I}$ is any indexed set of complex numbers, then the series $sum_{iin I}z_i$ is said to converge to the complex number $z$ if for every $epsilon>0$ there is a finite subset $J_epsilon$ of $I$ such that, for every finite subset $J$ of $I$ with $J_epsilonsubseteq J$, $vert sum_{iin J}z_i-zvert<epsilon$. In other words, to say that the series converges to $z$ is to say that the net $Jmapstosum_{iin J}z_i$, from finite subsets of $I$ directed by inclusion, converges to $z$. This is a special case of the standard definition of unordered summation which works in an arbitrary commutative (Hausdorff) topological group. A comprehensive reference for this kind of summation is section 5 of Chapter III of Bourbaki's General Topology. In particular, because $mathbf{C}$ is first-countable, the corollary on page 263 of loc. cit. implies that, if such a $z$ exists, then the set ${iin I:z_ineq 0}$ is countable (such a result was mentioned in the answer by Reese, but it is not necessary to use any kind of transfinite recursion to make sense of this kind of summation).



So, unless you want to introduce a nonstandard notion of summation (which necessarily must lack some of the features one would hope for based on the case of finite sums), the series in question cannot be meaningfully said to converge to anything.






share|cite|improve this answer































    0














    The obvious answer is that no such sum can be defined uniquely. If one limits to summation methods that are symmetric around the origin, the conditional limit could be zero. (The process would be to sum over squares (-x,-ix; x, -ix; x, ix, -x, ix) such that each set of 4 points sums to zero (as does the origin.) Of course,summing over (-x,-ix; 2x, -ix; 2x, ix, -x, ix) would give a result that becomes arbitrarily large. By playing around with summation methods one could generate any result.






    share|cite|improve this answer





















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      3 Answers
      3






      active

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      3 Answers
      3






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      active

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      active

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      19














      Traditionally, the sum of a sequence is defined as the limit of the partial sums; that is, for a sequence ${a_n}$, $sum{a_n}$ is that number $S$ so that for every $epsilon > 0$, there is an $N$ such that whenever $m > N$, $|S - sum_{n = 0}^ma_n| < epsilon$. There's no reason we can't define it like that for uncountable sequences as well: let $mathfrak{c}$ be the cardinality of $mathbb{C}$, and let ${a_{alpha}}$ be a sequence of complex numbers where the indices are ordinals less than $mathfrak{c}$. We define $sum{a_{alpha}}$ as that value $S$ so that for every $epsilon > 0$, there is a $beta < mathfrak{c}$ so that whenever $gamma > beta$, $|S - sum_{alpha = 0}^{gamma}a_{alpha}| < epsilon$. Note that this requires us to recursively define transfinite summation, to make sense of that sum up to $gamma$.



      But here's the thing: taking $epsilon$ to be $1$, then $1/2$, then $1/4$, and so on, we get a sequence of "threshold" $beta$ corresponding to each one; call $beta_n$ the $beta$ corresponding to $epsilon = 1/2^n$. This is a countable sequence (length strictly less than $mathfrak{c}$). Inconveniently, $mathfrak{c}$ is regular: any increasing sequence of ordinals less than $mathfrak{c}$ with length less than $mathfrak{c}$ must be bounded strictly below $mathfrak{c}$. So that means there's some $beta_{infty}$ that's below $mathfrak{c}$ but greater than every $beta_n$. But by definition, that means that all partial sums past $beta_{infty}$ are less than $1/2^n$ away from $S$ for every $n$. So they must be exactly equal to $S$. And that means that we must be only adding $0$ from that point forward.



      This is a well-known result that I can't recall a reference for: the only uncountable sequences that have convergent sums are those which consist of countably many nonzero terms followed by nothing but zeroes. In other words, there's no sensible way to sum over all of the complex numbers and get a convergence.






      share|cite|improve this answer

















      • 2




        How does one mentally reconcile this information with the idea of an integral? You could view that as a continuous sum.
        – orlp
        18 hours ago










      • That's a very good question, @orlp; I think I'll withdraw the acceptance of this answer until I understand what it implies about integrals.
        – Shaun
        18 hours ago






      • 1




        The regularity of $mathfrak{c}$ is really a red herring here (and in fact, $mathfrak{c}$ may not be regular!). You could just as well pick an enumeration of $mathbb{C}$ whose length has countable cofinality, and then there might not be any point beyond which all the terms are $0$. What you can prove though (independent of the ordering used) is that only countably many terms can be nonzero. (Proof: if there are uncountably many nonzero terms, apply your argument to the first partial sum which contains uncountably many nonzero terms, which must have cofinality $omega_1$.)
        – Eric Wofsey
        14 hours ago












      • As for the regularity of $mathfrak{c}$, it is consistent with ZFC that $mathfrak{c}$ is singular (for instance, it could be $aleph_{omega_1}$). However, it follows from König's theorem that it has uncountable cofiinality so your argument still does work for it.
        – Eric Wofsey
        14 hours ago






      • 3




        @orlp An integral isn't an infinite sum; it's an infinite average. You're effectively dividing each term by a measure of the "number" of terms, rendering it infinitesimal; in this case, the uncountable analogue of infinitesimal. That means it doesn't really resemble a "sum" in the sense we usually mean (for example, the integral of the function $f(x)$ which is $1$ at $0$ and $1$ and zero elsewhere is not $2$).
        – Reese
        9 hours ago
















      19














      Traditionally, the sum of a sequence is defined as the limit of the partial sums; that is, for a sequence ${a_n}$, $sum{a_n}$ is that number $S$ so that for every $epsilon > 0$, there is an $N$ such that whenever $m > N$, $|S - sum_{n = 0}^ma_n| < epsilon$. There's no reason we can't define it like that for uncountable sequences as well: let $mathfrak{c}$ be the cardinality of $mathbb{C}$, and let ${a_{alpha}}$ be a sequence of complex numbers where the indices are ordinals less than $mathfrak{c}$. We define $sum{a_{alpha}}$ as that value $S$ so that for every $epsilon > 0$, there is a $beta < mathfrak{c}$ so that whenever $gamma > beta$, $|S - sum_{alpha = 0}^{gamma}a_{alpha}| < epsilon$. Note that this requires us to recursively define transfinite summation, to make sense of that sum up to $gamma$.



      But here's the thing: taking $epsilon$ to be $1$, then $1/2$, then $1/4$, and so on, we get a sequence of "threshold" $beta$ corresponding to each one; call $beta_n$ the $beta$ corresponding to $epsilon = 1/2^n$. This is a countable sequence (length strictly less than $mathfrak{c}$). Inconveniently, $mathfrak{c}$ is regular: any increasing sequence of ordinals less than $mathfrak{c}$ with length less than $mathfrak{c}$ must be bounded strictly below $mathfrak{c}$. So that means there's some $beta_{infty}$ that's below $mathfrak{c}$ but greater than every $beta_n$. But by definition, that means that all partial sums past $beta_{infty}$ are less than $1/2^n$ away from $S$ for every $n$. So they must be exactly equal to $S$. And that means that we must be only adding $0$ from that point forward.



      This is a well-known result that I can't recall a reference for: the only uncountable sequences that have convergent sums are those which consist of countably many nonzero terms followed by nothing but zeroes. In other words, there's no sensible way to sum over all of the complex numbers and get a convergence.






      share|cite|improve this answer

















      • 2




        How does one mentally reconcile this information with the idea of an integral? You could view that as a continuous sum.
        – orlp
        18 hours ago










      • That's a very good question, @orlp; I think I'll withdraw the acceptance of this answer until I understand what it implies about integrals.
        – Shaun
        18 hours ago






      • 1




        The regularity of $mathfrak{c}$ is really a red herring here (and in fact, $mathfrak{c}$ may not be regular!). You could just as well pick an enumeration of $mathbb{C}$ whose length has countable cofinality, and then there might not be any point beyond which all the terms are $0$. What you can prove though (independent of the ordering used) is that only countably many terms can be nonzero. (Proof: if there are uncountably many nonzero terms, apply your argument to the first partial sum which contains uncountably many nonzero terms, which must have cofinality $omega_1$.)
        – Eric Wofsey
        14 hours ago












      • As for the regularity of $mathfrak{c}$, it is consistent with ZFC that $mathfrak{c}$ is singular (for instance, it could be $aleph_{omega_1}$). However, it follows from König's theorem that it has uncountable cofiinality so your argument still does work for it.
        – Eric Wofsey
        14 hours ago






      • 3




        @orlp An integral isn't an infinite sum; it's an infinite average. You're effectively dividing each term by a measure of the "number" of terms, rendering it infinitesimal; in this case, the uncountable analogue of infinitesimal. That means it doesn't really resemble a "sum" in the sense we usually mean (for example, the integral of the function $f(x)$ which is $1$ at $0$ and $1$ and zero elsewhere is not $2$).
        – Reese
        9 hours ago














      19












      19








      19






      Traditionally, the sum of a sequence is defined as the limit of the partial sums; that is, for a sequence ${a_n}$, $sum{a_n}$ is that number $S$ so that for every $epsilon > 0$, there is an $N$ such that whenever $m > N$, $|S - sum_{n = 0}^ma_n| < epsilon$. There's no reason we can't define it like that for uncountable sequences as well: let $mathfrak{c}$ be the cardinality of $mathbb{C}$, and let ${a_{alpha}}$ be a sequence of complex numbers where the indices are ordinals less than $mathfrak{c}$. We define $sum{a_{alpha}}$ as that value $S$ so that for every $epsilon > 0$, there is a $beta < mathfrak{c}$ so that whenever $gamma > beta$, $|S - sum_{alpha = 0}^{gamma}a_{alpha}| < epsilon$. Note that this requires us to recursively define transfinite summation, to make sense of that sum up to $gamma$.



      But here's the thing: taking $epsilon$ to be $1$, then $1/2$, then $1/4$, and so on, we get a sequence of "threshold" $beta$ corresponding to each one; call $beta_n$ the $beta$ corresponding to $epsilon = 1/2^n$. This is a countable sequence (length strictly less than $mathfrak{c}$). Inconveniently, $mathfrak{c}$ is regular: any increasing sequence of ordinals less than $mathfrak{c}$ with length less than $mathfrak{c}$ must be bounded strictly below $mathfrak{c}$. So that means there's some $beta_{infty}$ that's below $mathfrak{c}$ but greater than every $beta_n$. But by definition, that means that all partial sums past $beta_{infty}$ are less than $1/2^n$ away from $S$ for every $n$. So they must be exactly equal to $S$. And that means that we must be only adding $0$ from that point forward.



      This is a well-known result that I can't recall a reference for: the only uncountable sequences that have convergent sums are those which consist of countably many nonzero terms followed by nothing but zeroes. In other words, there's no sensible way to sum over all of the complex numbers and get a convergence.






      share|cite|improve this answer












      Traditionally, the sum of a sequence is defined as the limit of the partial sums; that is, for a sequence ${a_n}$, $sum{a_n}$ is that number $S$ so that for every $epsilon > 0$, there is an $N$ such that whenever $m > N$, $|S - sum_{n = 0}^ma_n| < epsilon$. There's no reason we can't define it like that for uncountable sequences as well: let $mathfrak{c}$ be the cardinality of $mathbb{C}$, and let ${a_{alpha}}$ be a sequence of complex numbers where the indices are ordinals less than $mathfrak{c}$. We define $sum{a_{alpha}}$ as that value $S$ so that for every $epsilon > 0$, there is a $beta < mathfrak{c}$ so that whenever $gamma > beta$, $|S - sum_{alpha = 0}^{gamma}a_{alpha}| < epsilon$. Note that this requires us to recursively define transfinite summation, to make sense of that sum up to $gamma$.



      But here's the thing: taking $epsilon$ to be $1$, then $1/2$, then $1/4$, and so on, we get a sequence of "threshold" $beta$ corresponding to each one; call $beta_n$ the $beta$ corresponding to $epsilon = 1/2^n$. This is a countable sequence (length strictly less than $mathfrak{c}$). Inconveniently, $mathfrak{c}$ is regular: any increasing sequence of ordinals less than $mathfrak{c}$ with length less than $mathfrak{c}$ must be bounded strictly below $mathfrak{c}$. So that means there's some $beta_{infty}$ that's below $mathfrak{c}$ but greater than every $beta_n$. But by definition, that means that all partial sums past $beta_{infty}$ are less than $1/2^n$ away from $S$ for every $n$. So they must be exactly equal to $S$. And that means that we must be only adding $0$ from that point forward.



      This is a well-known result that I can't recall a reference for: the only uncountable sequences that have convergent sums are those which consist of countably many nonzero terms followed by nothing but zeroes. In other words, there's no sensible way to sum over all of the complex numbers and get a convergence.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 21 hours ago









      Reese

      15.2k11238




      15.2k11238








      • 2




        How does one mentally reconcile this information with the idea of an integral? You could view that as a continuous sum.
        – orlp
        18 hours ago










      • That's a very good question, @orlp; I think I'll withdraw the acceptance of this answer until I understand what it implies about integrals.
        – Shaun
        18 hours ago






      • 1




        The regularity of $mathfrak{c}$ is really a red herring here (and in fact, $mathfrak{c}$ may not be regular!). You could just as well pick an enumeration of $mathbb{C}$ whose length has countable cofinality, and then there might not be any point beyond which all the terms are $0$. What you can prove though (independent of the ordering used) is that only countably many terms can be nonzero. (Proof: if there are uncountably many nonzero terms, apply your argument to the first partial sum which contains uncountably many nonzero terms, which must have cofinality $omega_1$.)
        – Eric Wofsey
        14 hours ago












      • As for the regularity of $mathfrak{c}$, it is consistent with ZFC that $mathfrak{c}$ is singular (for instance, it could be $aleph_{omega_1}$). However, it follows from König's theorem that it has uncountable cofiinality so your argument still does work for it.
        – Eric Wofsey
        14 hours ago






      • 3




        @orlp An integral isn't an infinite sum; it's an infinite average. You're effectively dividing each term by a measure of the "number" of terms, rendering it infinitesimal; in this case, the uncountable analogue of infinitesimal. That means it doesn't really resemble a "sum" in the sense we usually mean (for example, the integral of the function $f(x)$ which is $1$ at $0$ and $1$ and zero elsewhere is not $2$).
        – Reese
        9 hours ago














      • 2




        How does one mentally reconcile this information with the idea of an integral? You could view that as a continuous sum.
        – orlp
        18 hours ago










      • That's a very good question, @orlp; I think I'll withdraw the acceptance of this answer until I understand what it implies about integrals.
        – Shaun
        18 hours ago






      • 1




        The regularity of $mathfrak{c}$ is really a red herring here (and in fact, $mathfrak{c}$ may not be regular!). You could just as well pick an enumeration of $mathbb{C}$ whose length has countable cofinality, and then there might not be any point beyond which all the terms are $0$. What you can prove though (independent of the ordering used) is that only countably many terms can be nonzero. (Proof: if there are uncountably many nonzero terms, apply your argument to the first partial sum which contains uncountably many nonzero terms, which must have cofinality $omega_1$.)
        – Eric Wofsey
        14 hours ago












      • As for the regularity of $mathfrak{c}$, it is consistent with ZFC that $mathfrak{c}$ is singular (for instance, it could be $aleph_{omega_1}$). However, it follows from König's theorem that it has uncountable cofiinality so your argument still does work for it.
        – Eric Wofsey
        14 hours ago






      • 3




        @orlp An integral isn't an infinite sum; it's an infinite average. You're effectively dividing each term by a measure of the "number" of terms, rendering it infinitesimal; in this case, the uncountable analogue of infinitesimal. That means it doesn't really resemble a "sum" in the sense we usually mean (for example, the integral of the function $f(x)$ which is $1$ at $0$ and $1$ and zero elsewhere is not $2$).
        – Reese
        9 hours ago








      2




      2




      How does one mentally reconcile this information with the idea of an integral? You could view that as a continuous sum.
      – orlp
      18 hours ago




      How does one mentally reconcile this information with the idea of an integral? You could view that as a continuous sum.
      – orlp
      18 hours ago












      That's a very good question, @orlp; I think I'll withdraw the acceptance of this answer until I understand what it implies about integrals.
      – Shaun
      18 hours ago




      That's a very good question, @orlp; I think I'll withdraw the acceptance of this answer until I understand what it implies about integrals.
      – Shaun
      18 hours ago




      1




      1




      The regularity of $mathfrak{c}$ is really a red herring here (and in fact, $mathfrak{c}$ may not be regular!). You could just as well pick an enumeration of $mathbb{C}$ whose length has countable cofinality, and then there might not be any point beyond which all the terms are $0$. What you can prove though (independent of the ordering used) is that only countably many terms can be nonzero. (Proof: if there are uncountably many nonzero terms, apply your argument to the first partial sum which contains uncountably many nonzero terms, which must have cofinality $omega_1$.)
      – Eric Wofsey
      14 hours ago






      The regularity of $mathfrak{c}$ is really a red herring here (and in fact, $mathfrak{c}$ may not be regular!). You could just as well pick an enumeration of $mathbb{C}$ whose length has countable cofinality, and then there might not be any point beyond which all the terms are $0$. What you can prove though (independent of the ordering used) is that only countably many terms can be nonzero. (Proof: if there are uncountably many nonzero terms, apply your argument to the first partial sum which contains uncountably many nonzero terms, which must have cofinality $omega_1$.)
      – Eric Wofsey
      14 hours ago














      As for the regularity of $mathfrak{c}$, it is consistent with ZFC that $mathfrak{c}$ is singular (for instance, it could be $aleph_{omega_1}$). However, it follows from König's theorem that it has uncountable cofiinality so your argument still does work for it.
      – Eric Wofsey
      14 hours ago




      As for the regularity of $mathfrak{c}$, it is consistent with ZFC that $mathfrak{c}$ is singular (for instance, it could be $aleph_{omega_1}$). However, it follows from König's theorem that it has uncountable cofiinality so your argument still does work for it.
      – Eric Wofsey
      14 hours ago




      3




      3




      @orlp An integral isn't an infinite sum; it's an infinite average. You're effectively dividing each term by a measure of the "number" of terms, rendering it infinitesimal; in this case, the uncountable analogue of infinitesimal. That means it doesn't really resemble a "sum" in the sense we usually mean (for example, the integral of the function $f(x)$ which is $1$ at $0$ and $1$ and zero elsewhere is not $2$).
      – Reese
      9 hours ago




      @orlp An integral isn't an infinite sum; it's an infinite average. You're effectively dividing each term by a measure of the "number" of terms, rendering it infinitesimal; in this case, the uncountable analogue of infinitesimal. That means it doesn't really resemble a "sum" in the sense we usually mean (for example, the integral of the function $f(x)$ which is $1$ at $0$ and $1$ and zero elsewhere is not $2$).
      – Reese
      9 hours ago











      10














      If ${z_i:iin I}$ is any indexed set of complex numbers, then the series $sum_{iin I}z_i$ is said to converge to the complex number $z$ if for every $epsilon>0$ there is a finite subset $J_epsilon$ of $I$ such that, for every finite subset $J$ of $I$ with $J_epsilonsubseteq J$, $vert sum_{iin J}z_i-zvert<epsilon$. In other words, to say that the series converges to $z$ is to say that the net $Jmapstosum_{iin J}z_i$, from finite subsets of $I$ directed by inclusion, converges to $z$. This is a special case of the standard definition of unordered summation which works in an arbitrary commutative (Hausdorff) topological group. A comprehensive reference for this kind of summation is section 5 of Chapter III of Bourbaki's General Topology. In particular, because $mathbf{C}$ is first-countable, the corollary on page 263 of loc. cit. implies that, if such a $z$ exists, then the set ${iin I:z_ineq 0}$ is countable (such a result was mentioned in the answer by Reese, but it is not necessary to use any kind of transfinite recursion to make sense of this kind of summation).



      So, unless you want to introduce a nonstandard notion of summation (which necessarily must lack some of the features one would hope for based on the case of finite sums), the series in question cannot be meaningfully said to converge to anything.






      share|cite|improve this answer




























        10














        If ${z_i:iin I}$ is any indexed set of complex numbers, then the series $sum_{iin I}z_i$ is said to converge to the complex number $z$ if for every $epsilon>0$ there is a finite subset $J_epsilon$ of $I$ such that, for every finite subset $J$ of $I$ with $J_epsilonsubseteq J$, $vert sum_{iin J}z_i-zvert<epsilon$. In other words, to say that the series converges to $z$ is to say that the net $Jmapstosum_{iin J}z_i$, from finite subsets of $I$ directed by inclusion, converges to $z$. This is a special case of the standard definition of unordered summation which works in an arbitrary commutative (Hausdorff) topological group. A comprehensive reference for this kind of summation is section 5 of Chapter III of Bourbaki's General Topology. In particular, because $mathbf{C}$ is first-countable, the corollary on page 263 of loc. cit. implies that, if such a $z$ exists, then the set ${iin I:z_ineq 0}$ is countable (such a result was mentioned in the answer by Reese, but it is not necessary to use any kind of transfinite recursion to make sense of this kind of summation).



        So, unless you want to introduce a nonstandard notion of summation (which necessarily must lack some of the features one would hope for based on the case of finite sums), the series in question cannot be meaningfully said to converge to anything.






        share|cite|improve this answer


























          10












          10








          10






          If ${z_i:iin I}$ is any indexed set of complex numbers, then the series $sum_{iin I}z_i$ is said to converge to the complex number $z$ if for every $epsilon>0$ there is a finite subset $J_epsilon$ of $I$ such that, for every finite subset $J$ of $I$ with $J_epsilonsubseteq J$, $vert sum_{iin J}z_i-zvert<epsilon$. In other words, to say that the series converges to $z$ is to say that the net $Jmapstosum_{iin J}z_i$, from finite subsets of $I$ directed by inclusion, converges to $z$. This is a special case of the standard definition of unordered summation which works in an arbitrary commutative (Hausdorff) topological group. A comprehensive reference for this kind of summation is section 5 of Chapter III of Bourbaki's General Topology. In particular, because $mathbf{C}$ is first-countable, the corollary on page 263 of loc. cit. implies that, if such a $z$ exists, then the set ${iin I:z_ineq 0}$ is countable (such a result was mentioned in the answer by Reese, but it is not necessary to use any kind of transfinite recursion to make sense of this kind of summation).



          So, unless you want to introduce a nonstandard notion of summation (which necessarily must lack some of the features one would hope for based on the case of finite sums), the series in question cannot be meaningfully said to converge to anything.






          share|cite|improve this answer














          If ${z_i:iin I}$ is any indexed set of complex numbers, then the series $sum_{iin I}z_i$ is said to converge to the complex number $z$ if for every $epsilon>0$ there is a finite subset $J_epsilon$ of $I$ such that, for every finite subset $J$ of $I$ with $J_epsilonsubseteq J$, $vert sum_{iin J}z_i-zvert<epsilon$. In other words, to say that the series converges to $z$ is to say that the net $Jmapstosum_{iin J}z_i$, from finite subsets of $I$ directed by inclusion, converges to $z$. This is a special case of the standard definition of unordered summation which works in an arbitrary commutative (Hausdorff) topological group. A comprehensive reference for this kind of summation is section 5 of Chapter III of Bourbaki's General Topology. In particular, because $mathbf{C}$ is first-countable, the corollary on page 263 of loc. cit. implies that, if such a $z$ exists, then the set ${iin I:z_ineq 0}$ is countable (such a result was mentioned in the answer by Reese, but it is not necessary to use any kind of transfinite recursion to make sense of this kind of summation).



          So, unless you want to introduce a nonstandard notion of summation (which necessarily must lack some of the features one would hope for based on the case of finite sums), the series in question cannot be meaningfully said to converge to anything.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 19 hours ago

























          answered 20 hours ago









          Keenan Kidwell

          19.5k13472




          19.5k13472























              0














              The obvious answer is that no such sum can be defined uniquely. If one limits to summation methods that are symmetric around the origin, the conditional limit could be zero. (The process would be to sum over squares (-x,-ix; x, -ix; x, ix, -x, ix) such that each set of 4 points sums to zero (as does the origin.) Of course,summing over (-x,-ix; 2x, -ix; 2x, ix, -x, ix) would give a result that becomes arbitrarily large. By playing around with summation methods one could generate any result.






              share|cite|improve this answer


























                0














                The obvious answer is that no such sum can be defined uniquely. If one limits to summation methods that are symmetric around the origin, the conditional limit could be zero. (The process would be to sum over squares (-x,-ix; x, -ix; x, ix, -x, ix) such that each set of 4 points sums to zero (as does the origin.) Of course,summing over (-x,-ix; 2x, -ix; 2x, ix, -x, ix) would give a result that becomes arbitrarily large. By playing around with summation methods one could generate any result.






                share|cite|improve this answer
























                  0












                  0








                  0






                  The obvious answer is that no such sum can be defined uniquely. If one limits to summation methods that are symmetric around the origin, the conditional limit could be zero. (The process would be to sum over squares (-x,-ix; x, -ix; x, ix, -x, ix) such that each set of 4 points sums to zero (as does the origin.) Of course,summing over (-x,-ix; 2x, -ix; 2x, ix, -x, ix) would give a result that becomes arbitrarily large. By playing around with summation methods one could generate any result.






                  share|cite|improve this answer












                  The obvious answer is that no such sum can be defined uniquely. If one limits to summation methods that are symmetric around the origin, the conditional limit could be zero. (The process would be to sum over squares (-x,-ix; x, -ix; x, ix, -x, ix) such that each set of 4 points sums to zero (as does the origin.) Of course,summing over (-x,-ix; 2x, -ix; 2x, ix, -x, ix) would give a result that becomes arbitrarily large. By playing around with summation methods one could generate any result.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 16 hours ago









                  ttw

                  26124




                  26124






























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