Series convergence without sigma notation











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Consider the following series:



$$frac{1}{1} + frac{10}{2} + frac{100}{3} - frac{37}{4} - frac{37}{5} - frac{37}{6} + frac{1}{7} + frac{10}{8} + frac{100}{9} - frac{37}{10} - frac{37}{11} - frac{37}{12} + dots$$



This series seems to converge but I am not able to prove it. It seems impossible to write thing whole thing into summation notation. I tried to group by modulo 6, but then for example sum of $frac{1}{6n+1}$ diverges already so that doesn't seem right. Also tried to change the 37s into powers of 10 to maybe use the comparison test but also failed.
Any suggestions?










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  • Related: 1 and 2
    – Mason
    58 mins ago















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Consider the following series:



$$frac{1}{1} + frac{10}{2} + frac{100}{3} - frac{37}{4} - frac{37}{5} - frac{37}{6} + frac{1}{7} + frac{10}{8} + frac{100}{9} - frac{37}{10} - frac{37}{11} - frac{37}{12} + dots$$



This series seems to converge but I am not able to prove it. It seems impossible to write thing whole thing into summation notation. I tried to group by modulo 6, but then for example sum of $frac{1}{6n+1}$ diverges already so that doesn't seem right. Also tried to change the 37s into powers of 10 to maybe use the comparison test but also failed.
Any suggestions?










share|cite|improve this question






















  • Related: 1 and 2
    – Mason
    58 mins ago













up vote
6
down vote

favorite









up vote
6
down vote

favorite











Consider the following series:



$$frac{1}{1} + frac{10}{2} + frac{100}{3} - frac{37}{4} - frac{37}{5} - frac{37}{6} + frac{1}{7} + frac{10}{8} + frac{100}{9} - frac{37}{10} - frac{37}{11} - frac{37}{12} + dots$$



This series seems to converge but I am not able to prove it. It seems impossible to write thing whole thing into summation notation. I tried to group by modulo 6, but then for example sum of $frac{1}{6n+1}$ diverges already so that doesn't seem right. Also tried to change the 37s into powers of 10 to maybe use the comparison test but also failed.
Any suggestions?










share|cite|improve this question













Consider the following series:



$$frac{1}{1} + frac{10}{2} + frac{100}{3} - frac{37}{4} - frac{37}{5} - frac{37}{6} + frac{1}{7} + frac{10}{8} + frac{100}{9} - frac{37}{10} - frac{37}{11} - frac{37}{12} + dots$$



This series seems to converge but I am not able to prove it. It seems impossible to write thing whole thing into summation notation. I tried to group by modulo 6, but then for example sum of $frac{1}{6n+1}$ diverges already so that doesn't seem right. Also tried to change the 37s into powers of 10 to maybe use the comparison test but also failed.
Any suggestions?







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asked 5 hours ago









Markus Punnar

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  • Related: 1 and 2
    – Mason
    58 mins ago


















  • Related: 1 and 2
    – Mason
    58 mins ago
















Related: 1 and 2
– Mason
58 mins ago




Related: 1 and 2
– Mason
58 mins ago










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First, notice that $$F_n=frac{1}{6n+1}+frac{10}{6n+2}+frac{100}{6n+3}-frac{37}{6n+4}-frac{37}{6n+5}-frac{37}{6n+6} geq frac{111}{6n+3}-frac{111}{6n+4} > 0.$$
Then, notice that $$F_n leq frac{111}{6n+1}-frac{111}{6n+6}=frac{555}{(6n+1)(6n+6)}.$$



Thus the sum of $F_n$ converges. How can you infer the final result from this?






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  • Alternatively, you can write $$F_n = {frac {303264,{n}^{4}+649296,{n}^{3}+471096,{n}^{2}+132516,n+ 11892}{ left( 6,n+1 right) left( 6,n+2 right) left( 6,n+3 right) left( 6,n+4 right) left( 6,n+5 right) left( 6,n+6 right) }} $$ and note that the numerator has degree $4$ and denominator degree $6$, so $F_n = O(1/n^2)$.
    – Robert Israel
    4 hours ago


















up vote
4
down vote













I presume the pattern $(1, 10, 100, -37, -37, -37)$ continues forever, so this is
$$ sum_{n=1}^infty frac{a_n}{n}$$
where $$ a_n = cases{1 & if $n equiv 1 mod 6$cr
10 & if $n equiv 2 mod 6$cr
100 & if $n equiv 3 mod 6$cr
-37 & if $n equiv 4,5$ or $0 mod 6$cr}$$

Note that $1+10+100 - 3times 37 = 0$. Then the $6m$
-th partial sum
$$ eqalign{S_m &= sum_{n=1}^{6m} frac{a_n}{n}cr
& = sum_{j=0}^m frac{1}{6j+1} +
10 sum_{j=0}^m frac{1}{6j+2} + 100 sum_{j=0}^m frac{1}{6j+3}
- 37 sum_{k=4}^6 sum_{j=0}^m frac{1}{6j+k}cr
&= frac{1}{6} left(Psi(m+1+1/6) - Psi(1/6) + 10 (Psi(m+1+2/6) - Psi(2/6)) + 100 (Psi(m+1+3/6) - Psi(3/6)) - 37 (Psi(m+1+4/6)+Psi(m+1+5/6)+Psi(m+1+6/6)-Psi(4/6)-Psi(5/6)-Psi(6/6))right)cr}$$



and since $Psi(x) = ln(x) + O(1/x)$ as $x to infty$, the sum converges.
In fact, the limit is
$$ frac{64}{3} ln(2) - frac{63}{4} ln(3) +frac{161}{36} pi sqrt{3}$$






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    up vote
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    First, notice that $$F_n=frac{1}{6n+1}+frac{10}{6n+2}+frac{100}{6n+3}-frac{37}{6n+4}-frac{37}{6n+5}-frac{37}{6n+6} geq frac{111}{6n+3}-frac{111}{6n+4} > 0.$$
    Then, notice that $$F_n leq frac{111}{6n+1}-frac{111}{6n+6}=frac{555}{(6n+1)(6n+6)}.$$



    Thus the sum of $F_n$ converges. How can you infer the final result from this?






    share|cite|improve this answer








    New contributor




    Mindlack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.


















    • Alternatively, you can write $$F_n = {frac {303264,{n}^{4}+649296,{n}^{3}+471096,{n}^{2}+132516,n+ 11892}{ left( 6,n+1 right) left( 6,n+2 right) left( 6,n+3 right) left( 6,n+4 right) left( 6,n+5 right) left( 6,n+6 right) }} $$ and note that the numerator has degree $4$ and denominator degree $6$, so $F_n = O(1/n^2)$.
      – Robert Israel
      4 hours ago















    up vote
    6
    down vote













    First, notice that $$F_n=frac{1}{6n+1}+frac{10}{6n+2}+frac{100}{6n+3}-frac{37}{6n+4}-frac{37}{6n+5}-frac{37}{6n+6} geq frac{111}{6n+3}-frac{111}{6n+4} > 0.$$
    Then, notice that $$F_n leq frac{111}{6n+1}-frac{111}{6n+6}=frac{555}{(6n+1)(6n+6)}.$$



    Thus the sum of $F_n$ converges. How can you infer the final result from this?






    share|cite|improve this answer








    New contributor




    Mindlack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.


















    • Alternatively, you can write $$F_n = {frac {303264,{n}^{4}+649296,{n}^{3}+471096,{n}^{2}+132516,n+ 11892}{ left( 6,n+1 right) left( 6,n+2 right) left( 6,n+3 right) left( 6,n+4 right) left( 6,n+5 right) left( 6,n+6 right) }} $$ and note that the numerator has degree $4$ and denominator degree $6$, so $F_n = O(1/n^2)$.
      – Robert Israel
      4 hours ago













    up vote
    6
    down vote










    up vote
    6
    down vote









    First, notice that $$F_n=frac{1}{6n+1}+frac{10}{6n+2}+frac{100}{6n+3}-frac{37}{6n+4}-frac{37}{6n+5}-frac{37}{6n+6} geq frac{111}{6n+3}-frac{111}{6n+4} > 0.$$
    Then, notice that $$F_n leq frac{111}{6n+1}-frac{111}{6n+6}=frac{555}{(6n+1)(6n+6)}.$$



    Thus the sum of $F_n$ converges. How can you infer the final result from this?






    share|cite|improve this answer








    New contributor




    Mindlack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    First, notice that $$F_n=frac{1}{6n+1}+frac{10}{6n+2}+frac{100}{6n+3}-frac{37}{6n+4}-frac{37}{6n+5}-frac{37}{6n+6} geq frac{111}{6n+3}-frac{111}{6n+4} > 0.$$
    Then, notice that $$F_n leq frac{111}{6n+1}-frac{111}{6n+6}=frac{555}{(6n+1)(6n+6)}.$$



    Thus the sum of $F_n$ converges. How can you infer the final result from this?







    share|cite|improve this answer








    New contributor




    Mindlack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    share|cite|improve this answer



    share|cite|improve this answer






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    answered 5 hours ago









    Mindlack

    4094




    4094




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    New contributor





    Mindlack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    Mindlack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.












    • Alternatively, you can write $$F_n = {frac {303264,{n}^{4}+649296,{n}^{3}+471096,{n}^{2}+132516,n+ 11892}{ left( 6,n+1 right) left( 6,n+2 right) left( 6,n+3 right) left( 6,n+4 right) left( 6,n+5 right) left( 6,n+6 right) }} $$ and note that the numerator has degree $4$ and denominator degree $6$, so $F_n = O(1/n^2)$.
      – Robert Israel
      4 hours ago


















    • Alternatively, you can write $$F_n = {frac {303264,{n}^{4}+649296,{n}^{3}+471096,{n}^{2}+132516,n+ 11892}{ left( 6,n+1 right) left( 6,n+2 right) left( 6,n+3 right) left( 6,n+4 right) left( 6,n+5 right) left( 6,n+6 right) }} $$ and note that the numerator has degree $4$ and denominator degree $6$, so $F_n = O(1/n^2)$.
      – Robert Israel
      4 hours ago
















    Alternatively, you can write $$F_n = {frac {303264,{n}^{4}+649296,{n}^{3}+471096,{n}^{2}+132516,n+ 11892}{ left( 6,n+1 right) left( 6,n+2 right) left( 6,n+3 right) left( 6,n+4 right) left( 6,n+5 right) left( 6,n+6 right) }} $$ and note that the numerator has degree $4$ and denominator degree $6$, so $F_n = O(1/n^2)$.
    – Robert Israel
    4 hours ago




    Alternatively, you can write $$F_n = {frac {303264,{n}^{4}+649296,{n}^{3}+471096,{n}^{2}+132516,n+ 11892}{ left( 6,n+1 right) left( 6,n+2 right) left( 6,n+3 right) left( 6,n+4 right) left( 6,n+5 right) left( 6,n+6 right) }} $$ and note that the numerator has degree $4$ and denominator degree $6$, so $F_n = O(1/n^2)$.
    – Robert Israel
    4 hours ago










    up vote
    4
    down vote













    I presume the pattern $(1, 10, 100, -37, -37, -37)$ continues forever, so this is
    $$ sum_{n=1}^infty frac{a_n}{n}$$
    where $$ a_n = cases{1 & if $n equiv 1 mod 6$cr
    10 & if $n equiv 2 mod 6$cr
    100 & if $n equiv 3 mod 6$cr
    -37 & if $n equiv 4,5$ or $0 mod 6$cr}$$

    Note that $1+10+100 - 3times 37 = 0$. Then the $6m$
    -th partial sum
    $$ eqalign{S_m &= sum_{n=1}^{6m} frac{a_n}{n}cr
    & = sum_{j=0}^m frac{1}{6j+1} +
    10 sum_{j=0}^m frac{1}{6j+2} + 100 sum_{j=0}^m frac{1}{6j+3}
    - 37 sum_{k=4}^6 sum_{j=0}^m frac{1}{6j+k}cr
    &= frac{1}{6} left(Psi(m+1+1/6) - Psi(1/6) + 10 (Psi(m+1+2/6) - Psi(2/6)) + 100 (Psi(m+1+3/6) - Psi(3/6)) - 37 (Psi(m+1+4/6)+Psi(m+1+5/6)+Psi(m+1+6/6)-Psi(4/6)-Psi(5/6)-Psi(6/6))right)cr}$$



    and since $Psi(x) = ln(x) + O(1/x)$ as $x to infty$, the sum converges.
    In fact, the limit is
    $$ frac{64}{3} ln(2) - frac{63}{4} ln(3) +frac{161}{36} pi sqrt{3}$$






    share|cite|improve this answer

























      up vote
      4
      down vote













      I presume the pattern $(1, 10, 100, -37, -37, -37)$ continues forever, so this is
      $$ sum_{n=1}^infty frac{a_n}{n}$$
      where $$ a_n = cases{1 & if $n equiv 1 mod 6$cr
      10 & if $n equiv 2 mod 6$cr
      100 & if $n equiv 3 mod 6$cr
      -37 & if $n equiv 4,5$ or $0 mod 6$cr}$$

      Note that $1+10+100 - 3times 37 = 0$. Then the $6m$
      -th partial sum
      $$ eqalign{S_m &= sum_{n=1}^{6m} frac{a_n}{n}cr
      & = sum_{j=0}^m frac{1}{6j+1} +
      10 sum_{j=0}^m frac{1}{6j+2} + 100 sum_{j=0}^m frac{1}{6j+3}
      - 37 sum_{k=4}^6 sum_{j=0}^m frac{1}{6j+k}cr
      &= frac{1}{6} left(Psi(m+1+1/6) - Psi(1/6) + 10 (Psi(m+1+2/6) - Psi(2/6)) + 100 (Psi(m+1+3/6) - Psi(3/6)) - 37 (Psi(m+1+4/6)+Psi(m+1+5/6)+Psi(m+1+6/6)-Psi(4/6)-Psi(5/6)-Psi(6/6))right)cr}$$



      and since $Psi(x) = ln(x) + O(1/x)$ as $x to infty$, the sum converges.
      In fact, the limit is
      $$ frac{64}{3} ln(2) - frac{63}{4} ln(3) +frac{161}{36} pi sqrt{3}$$






      share|cite|improve this answer























        up vote
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        down vote










        up vote
        4
        down vote









        I presume the pattern $(1, 10, 100, -37, -37, -37)$ continues forever, so this is
        $$ sum_{n=1}^infty frac{a_n}{n}$$
        where $$ a_n = cases{1 & if $n equiv 1 mod 6$cr
        10 & if $n equiv 2 mod 6$cr
        100 & if $n equiv 3 mod 6$cr
        -37 & if $n equiv 4,5$ or $0 mod 6$cr}$$

        Note that $1+10+100 - 3times 37 = 0$. Then the $6m$
        -th partial sum
        $$ eqalign{S_m &= sum_{n=1}^{6m} frac{a_n}{n}cr
        & = sum_{j=0}^m frac{1}{6j+1} +
        10 sum_{j=0}^m frac{1}{6j+2} + 100 sum_{j=0}^m frac{1}{6j+3}
        - 37 sum_{k=4}^6 sum_{j=0}^m frac{1}{6j+k}cr
        &= frac{1}{6} left(Psi(m+1+1/6) - Psi(1/6) + 10 (Psi(m+1+2/6) - Psi(2/6)) + 100 (Psi(m+1+3/6) - Psi(3/6)) - 37 (Psi(m+1+4/6)+Psi(m+1+5/6)+Psi(m+1+6/6)-Psi(4/6)-Psi(5/6)-Psi(6/6))right)cr}$$



        and since $Psi(x) = ln(x) + O(1/x)$ as $x to infty$, the sum converges.
        In fact, the limit is
        $$ frac{64}{3} ln(2) - frac{63}{4} ln(3) +frac{161}{36} pi sqrt{3}$$






        share|cite|improve this answer












        I presume the pattern $(1, 10, 100, -37, -37, -37)$ continues forever, so this is
        $$ sum_{n=1}^infty frac{a_n}{n}$$
        where $$ a_n = cases{1 & if $n equiv 1 mod 6$cr
        10 & if $n equiv 2 mod 6$cr
        100 & if $n equiv 3 mod 6$cr
        -37 & if $n equiv 4,5$ or $0 mod 6$cr}$$

        Note that $1+10+100 - 3times 37 = 0$. Then the $6m$
        -th partial sum
        $$ eqalign{S_m &= sum_{n=1}^{6m} frac{a_n}{n}cr
        & = sum_{j=0}^m frac{1}{6j+1} +
        10 sum_{j=0}^m frac{1}{6j+2} + 100 sum_{j=0}^m frac{1}{6j+3}
        - 37 sum_{k=4}^6 sum_{j=0}^m frac{1}{6j+k}cr
        &= frac{1}{6} left(Psi(m+1+1/6) - Psi(1/6) + 10 (Psi(m+1+2/6) - Psi(2/6)) + 100 (Psi(m+1+3/6) - Psi(3/6)) - 37 (Psi(m+1+4/6)+Psi(m+1+5/6)+Psi(m+1+6/6)-Psi(4/6)-Psi(5/6)-Psi(6/6))right)cr}$$



        and since $Psi(x) = ln(x) + O(1/x)$ as $x to infty$, the sum converges.
        In fact, the limit is
        $$ frac{64}{3} ln(2) - frac{63}{4} ln(3) +frac{161}{36} pi sqrt{3}$$







        share|cite|improve this answer












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        answered 5 hours ago









        Robert Israel

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