An alternative way to find the sum of this series? [duplicate]












6












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This question already has an answer here:




  • Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $

    7 answers




$displaystyle frac{4}{20}$+$displaystyle frac{4.7}{20.30}$+$displaystyle frac{4.7.10}{20.30.40}$+...



Now I have tried to solve this in a usual way, first find the nth term $t_n$.



$t_n$= $displaystyle frac{1}{10}$($displaystyle frac{1+3}{2}$) + $displaystyle frac{1}{10^2}$($displaystyle frac{1+3}{2}$)($displaystyle frac{1+6}{3}$) + ...+ $displaystyle frac{1}{10^n}$($displaystyle frac{1+3}{2}$)($displaystyle frac{1+6}{3}$)...($displaystyle frac{1+3n}{n+1}$)



=$displaystyle frac{1}{10^n}prod$(1+$displaystyle frac{2r}{r+1}$) , $r=1,2,..,n$



=$displaystyle prod$($displaystyle frac{3}{10}-displaystyle frac{1}{5(r+1)}$)
thus, $t_n=$ (x-$displaystyle frac{a}{2}$)(x-$displaystyle frac{a}{3}$)...(x-$displaystyle frac{a}{n+1}$), x=$displaystyle frac{3}{10}$, a=$displaystyle frac{1}{5}$



Now to calculate $S_n$, I have to find the product $t_n$, and then take sum over it. But this seems to be a very tedious job. Is there any elegant method(may be using the expansions of any analytic functions) to do this?






real-analysis sequences-and-series







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 Jan 22 at 9:04


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    This is just binomial series.
    $endgroup$
    – Nemo
    Jan 19 '18 at 17:19
















6












$begingroup$



This question already has an answer here:




  • Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $

    7 answers




$displaystyle frac{4}{20}$+$displaystyle frac{4.7}{20.30}$+$displaystyle frac{4.7.10}{20.30.40}$+...



Now I have tried to solve this in a usual way, first find the nth term $t_n$.



$t_n$= $displaystyle frac{1}{10}$($displaystyle frac{1+3}{2}$) + $displaystyle frac{1}{10^2}$($displaystyle frac{1+3}{2}$)($displaystyle frac{1+6}{3}$) + ...+ $displaystyle frac{1}{10^n}$($displaystyle frac{1+3}{2}$)($displaystyle frac{1+6}{3}$)...($displaystyle frac{1+3n}{n+1}$)



=$displaystyle frac{1}{10^n}prod$(1+$displaystyle frac{2r}{r+1}$) , $r=1,2,..,n$



=$displaystyle prod$($displaystyle frac{3}{10}-displaystyle frac{1}{5(r+1)}$)
thus, $t_n=$ (x-$displaystyle frac{a}{2}$)(x-$displaystyle frac{a}{3}$)...(x-$displaystyle frac{a}{n+1}$), x=$displaystyle frac{3}{10}$, a=$displaystyle frac{1}{5}$



Now to calculate $S_n$, I have to find the product $t_n$, and then take sum over it. But this seems to be a very tedious job. Is there any elegant method(may be using the expansions of any analytic functions) to do this?






real-analysis sequences-and-series







share|cite|improve this question











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 Jan 22 at 9:04


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    This is just binomial series.
    $endgroup$
    – Nemo
    Jan 19 '18 at 17:19














6












6








6


1



$begingroup$



This question already has an answer here:




  • Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $

    7 answers




$displaystyle frac{4}{20}$+$displaystyle frac{4.7}{20.30}$+$displaystyle frac{4.7.10}{20.30.40}$+...



Now I have tried to solve this in a usual way, first find the nth term $t_n$.



$t_n$= $displaystyle frac{1}{10}$($displaystyle frac{1+3}{2}$) + $displaystyle frac{1}{10^2}$($displaystyle frac{1+3}{2}$)($displaystyle frac{1+6}{3}$) + ...+ $displaystyle frac{1}{10^n}$($displaystyle frac{1+3}{2}$)($displaystyle frac{1+6}{3}$)...($displaystyle frac{1+3n}{n+1}$)



=$displaystyle frac{1}{10^n}prod$(1+$displaystyle frac{2r}{r+1}$) , $r=1,2,..,n$



=$displaystyle prod$($displaystyle frac{3}{10}-displaystyle frac{1}{5(r+1)}$)
thus, $t_n=$ (x-$displaystyle frac{a}{2}$)(x-$displaystyle frac{a}{3}$)...(x-$displaystyle frac{a}{n+1}$), x=$displaystyle frac{3}{10}$, a=$displaystyle frac{1}{5}$



Now to calculate $S_n$, I have to find the product $t_n$, and then take sum over it. But this seems to be a very tedious job. Is there any elegant method(may be using the expansions of any analytic functions) to do this?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $

    7 answers




$displaystyle frac{4}{20}$+$displaystyle frac{4.7}{20.30}$+$displaystyle frac{4.7.10}{20.30.40}$+...



Now I have tried to solve this in a usual way, first find the nth term $t_n$.



$t_n$= $displaystyle frac{1}{10}$($displaystyle frac{1+3}{2}$) + $displaystyle frac{1}{10^2}$($displaystyle frac{1+3}{2}$)($displaystyle frac{1+6}{3}$) + ...+ $displaystyle frac{1}{10^n}$($displaystyle frac{1+3}{2}$)($displaystyle frac{1+6}{3}$)...($displaystyle frac{1+3n}{n+1}$)



=$displaystyle frac{1}{10^n}prod$(1+$displaystyle frac{2r}{r+1}$) , $r=1,2,..,n$



=$displaystyle prod$($displaystyle frac{3}{10}-displaystyle frac{1}{5(r+1)}$)
thus, $t_n=$ (x-$displaystyle frac{a}{2}$)(x-$displaystyle frac{a}{3}$)...(x-$displaystyle frac{a}{n+1}$), x=$displaystyle frac{3}{10}$, a=$displaystyle frac{1}{5}$



Now to calculate $S_n$, I have to find the product $t_n$, and then take sum over it. But this seems to be a very tedious job. Is there any elegant method(may be using the expansions of any analytic functions) to do this?





This question already has an answer here:




  • Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $

    7 answers







real-analysis sequences-and-series




real-analysis sequences-and-series






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edited Jan 18 '18 at 23:22









Mephlip

2121211




2121211










asked Jan 18 '18 at 21:17









Rio DuttaRio Dutta

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marked as duplicate by lab bhattacharjee sequences-and-series
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 Jan 22 at 9:04


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marked as duplicate by lab bhattacharjee sequences-and-series
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 Jan 22 at 9:04


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    This is just binomial series.
    $endgroup$
    – Nemo
    Jan 19 '18 at 17:19


















  • $begingroup$
    This is just binomial series.
    $endgroup$
    – Nemo
    Jan 19 '18 at 17:19
















$begingroup$
This is just binomial series.
$endgroup$
– Nemo
Jan 19 '18 at 17:19




$begingroup$
This is just binomial series.
$endgroup$
– Nemo
Jan 19 '18 at 17:19










2 Answers
2






active

oldest

votes


















6












$begingroup$

Through Euler's Beta function and the reflection formula for the $Gamma$ function:
$$sum_{ngeq 1}frac{prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=sum_{ngeq 1}frac{3^nGammaleft(n+frac{4}{3}right)}{10^n Gamma(n+2)Gammaleft(frac{4}{3}right)}=frac{3sqrt{3}}{2pi}sum_{ngeq 1}left(tfrac{3}{10}right)^n Bleft(tfrac{2}{3},n+tfrac{4}{3}right) $$
where
$$ sum_{ngeq 1}left(tfrac{3}{10}right)^n Bleft(tfrac{2}{3},n+tfrac{4}{3}right) = int_{0}^{1}sum_{ngeq 1}left(tfrac{3}{10}right)^n(1-x)^{-1/3}x^{n+1/3},dx=int_{0}^{1}frac{3x^{4/3},dx}{(1-x)^{1/3}(10-3x)} $$
and the last integral can be computed in a explicit way with a bit of patience. The final outcome is
$$sum_{ngeq 1}frac{prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=color{red}{10sqrt[3]{frac{10}{7}}-11} $$
which can also be proved by invoking Lagrange's inversion theorem or the extended binomial theorem.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
    newcommand{dd}{mathrm{d}}
    newcommand{ds}[1]{displaystyle{#1}}
    newcommand{expo}[1]{,mathrm{e}^{#1},}
    newcommand{ic}{mathrm{i}}
    newcommand{mc}[1]{mathcal{#1}}
    newcommand{mrm}[1]{mathrm{#1}}
    newcommand{pars}[1]{left(,{#1},right)}
    newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
    newcommand{root}[2]{,sqrt[#1]{,{#2},},}
    newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
    newcommand{verts}[1]{leftvert,{#1},rightvert}$
    begin{align}
    &bbox[10px,#ffd]{sum_{n = 1}^{infty}{prod_{k = 1}^{n}pars{3k + 1} over 10^{n}pars{n + 1}!}} =
    sum_{n = 2}^{infty}{3^{n - 1}
    prod_{k = 1}^{n - 1}pars{k + 1/3} over 10^{n - 1},n!}
    \[5mm] = &
    {10 over 3}sum_{n = 2}^{infty}pars{3 over 10}^{n},
    {Gammapars{4/3 + bracks{n - 1}}/Gammapars{4/3} over n!}
    \[5mm] = &
    {10 over 3},{pars{-2/3}! over Gammapars{4/3}}
    sum_{n = 2}^{infty}pars{3 over 10}^{n},
    {pars{n - 2/3}! over n!pars{-2/3}!}
    \[5mm] = &
    {10 over 3},{Gammapars{1/3} over pars{1/3}Gammapars{1/3}}
    sum_{n = 2}^{infty}pars{3 over 10}^{n},{n - 2/3 choose n}
    \[5mm] = &
    10sum_{n = 2}^{infty}pars{3 over 10}^{n}
    bracks{{-1/3 choose n}pars{-1}^{n}}
    \[5mm] = &
    10bracks{%
    sum_{n = 0}^{infty}{-1/3 choose n}pars{-,{3 over 10}}^{n}
    - overbrace{-1/3 choose 0}^{ds{= 1}} -
    overbrace{-1/3 choose 1}^{ds{= -,{1 over 3}}}
    pars{-,{3 over 10}}}
    \[5mm] = &
    10braces{bracks{1 + pars{-,{3 over 10}}}^{-1/3} - 1 - {1 over 10}}
    =
    bbx{10pars{10 over 7}^{1/3} - 11}
    \[5mm] approx & 0.2625
    end{align}






    share|cite|improve this answer











    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      Through Euler's Beta function and the reflection formula for the $Gamma$ function:
      $$sum_{ngeq 1}frac{prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=sum_{ngeq 1}frac{3^nGammaleft(n+frac{4}{3}right)}{10^n Gamma(n+2)Gammaleft(frac{4}{3}right)}=frac{3sqrt{3}}{2pi}sum_{ngeq 1}left(tfrac{3}{10}right)^n Bleft(tfrac{2}{3},n+tfrac{4}{3}right) $$
      where
      $$ sum_{ngeq 1}left(tfrac{3}{10}right)^n Bleft(tfrac{2}{3},n+tfrac{4}{3}right) = int_{0}^{1}sum_{ngeq 1}left(tfrac{3}{10}right)^n(1-x)^{-1/3}x^{n+1/3},dx=int_{0}^{1}frac{3x^{4/3},dx}{(1-x)^{1/3}(10-3x)} $$
      and the last integral can be computed in a explicit way with a bit of patience. The final outcome is
      $$sum_{ngeq 1}frac{prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=color{red}{10sqrt[3]{frac{10}{7}}-11} $$
      which can also be proved by invoking Lagrange's inversion theorem or the extended binomial theorem.






      share|cite|improve this answer









      $endgroup$


















        6












        $begingroup$

        Through Euler's Beta function and the reflection formula for the $Gamma$ function:
        $$sum_{ngeq 1}frac{prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=sum_{ngeq 1}frac{3^nGammaleft(n+frac{4}{3}right)}{10^n Gamma(n+2)Gammaleft(frac{4}{3}right)}=frac{3sqrt{3}}{2pi}sum_{ngeq 1}left(tfrac{3}{10}right)^n Bleft(tfrac{2}{3},n+tfrac{4}{3}right) $$
        where
        $$ sum_{ngeq 1}left(tfrac{3}{10}right)^n Bleft(tfrac{2}{3},n+tfrac{4}{3}right) = int_{0}^{1}sum_{ngeq 1}left(tfrac{3}{10}right)^n(1-x)^{-1/3}x^{n+1/3},dx=int_{0}^{1}frac{3x^{4/3},dx}{(1-x)^{1/3}(10-3x)} $$
        and the last integral can be computed in a explicit way with a bit of patience. The final outcome is
        $$sum_{ngeq 1}frac{prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=color{red}{10sqrt[3]{frac{10}{7}}-11} $$
        which can also be proved by invoking Lagrange's inversion theorem or the extended binomial theorem.






        share|cite|improve this answer









        $endgroup$
















          6












          6








          6





          $begingroup$

          Through Euler's Beta function and the reflection formula for the $Gamma$ function:
          $$sum_{ngeq 1}frac{prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=sum_{ngeq 1}frac{3^nGammaleft(n+frac{4}{3}right)}{10^n Gamma(n+2)Gammaleft(frac{4}{3}right)}=frac{3sqrt{3}}{2pi}sum_{ngeq 1}left(tfrac{3}{10}right)^n Bleft(tfrac{2}{3},n+tfrac{4}{3}right) $$
          where
          $$ sum_{ngeq 1}left(tfrac{3}{10}right)^n Bleft(tfrac{2}{3},n+tfrac{4}{3}right) = int_{0}^{1}sum_{ngeq 1}left(tfrac{3}{10}right)^n(1-x)^{-1/3}x^{n+1/3},dx=int_{0}^{1}frac{3x^{4/3},dx}{(1-x)^{1/3}(10-3x)} $$
          and the last integral can be computed in a explicit way with a bit of patience. The final outcome is
          $$sum_{ngeq 1}frac{prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=color{red}{10sqrt[3]{frac{10}{7}}-11} $$
          which can also be proved by invoking Lagrange's inversion theorem or the extended binomial theorem.






          share|cite|improve this answer









          $endgroup$



          Through Euler's Beta function and the reflection formula for the $Gamma$ function:
          $$sum_{ngeq 1}frac{prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=sum_{ngeq 1}frac{3^nGammaleft(n+frac{4}{3}right)}{10^n Gamma(n+2)Gammaleft(frac{4}{3}right)}=frac{3sqrt{3}}{2pi}sum_{ngeq 1}left(tfrac{3}{10}right)^n Bleft(tfrac{2}{3},n+tfrac{4}{3}right) $$
          where
          $$ sum_{ngeq 1}left(tfrac{3}{10}right)^n Bleft(tfrac{2}{3},n+tfrac{4}{3}right) = int_{0}^{1}sum_{ngeq 1}left(tfrac{3}{10}right)^n(1-x)^{-1/3}x^{n+1/3},dx=int_{0}^{1}frac{3x^{4/3},dx}{(1-x)^{1/3}(10-3x)} $$
          and the last integral can be computed in a explicit way with a bit of patience. The final outcome is
          $$sum_{ngeq 1}frac{prod_{k=1}^{n}(3k+1)}{10^n(n+1)!}=color{red}{10sqrt[3]{frac{10}{7}}-11} $$
          which can also be proved by invoking Lagrange's inversion theorem or the extended binomial theorem.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 18 '18 at 21:29









          Jack D'AurizioJack D'Aurizio

          291k33284669




          291k33284669























              2












              $begingroup$

              $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
              newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
              newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
              newcommand{dd}{mathrm{d}}
              newcommand{ds}[1]{displaystyle{#1}}
              newcommand{expo}[1]{,mathrm{e}^{#1},}
              newcommand{ic}{mathrm{i}}
              newcommand{mc}[1]{mathcal{#1}}
              newcommand{mrm}[1]{mathrm{#1}}
              newcommand{pars}[1]{left(,{#1},right)}
              newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
              newcommand{root}[2]{,sqrt[#1]{,{#2},},}
              newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
              newcommand{verts}[1]{leftvert,{#1},rightvert}$
              begin{align}
              &bbox[10px,#ffd]{sum_{n = 1}^{infty}{prod_{k = 1}^{n}pars{3k + 1} over 10^{n}pars{n + 1}!}} =
              sum_{n = 2}^{infty}{3^{n - 1}
              prod_{k = 1}^{n - 1}pars{k + 1/3} over 10^{n - 1},n!}
              \[5mm] = &
              {10 over 3}sum_{n = 2}^{infty}pars{3 over 10}^{n},
              {Gammapars{4/3 + bracks{n - 1}}/Gammapars{4/3} over n!}
              \[5mm] = &
              {10 over 3},{pars{-2/3}! over Gammapars{4/3}}
              sum_{n = 2}^{infty}pars{3 over 10}^{n},
              {pars{n - 2/3}! over n!pars{-2/3}!}
              \[5mm] = &
              {10 over 3},{Gammapars{1/3} over pars{1/3}Gammapars{1/3}}
              sum_{n = 2}^{infty}pars{3 over 10}^{n},{n - 2/3 choose n}
              \[5mm] = &
              10sum_{n = 2}^{infty}pars{3 over 10}^{n}
              bracks{{-1/3 choose n}pars{-1}^{n}}
              \[5mm] = &
              10bracks{%
              sum_{n = 0}^{infty}{-1/3 choose n}pars{-,{3 over 10}}^{n}
              - overbrace{-1/3 choose 0}^{ds{= 1}} -
              overbrace{-1/3 choose 1}^{ds{= -,{1 over 3}}}
              pars{-,{3 over 10}}}
              \[5mm] = &
              10braces{bracks{1 + pars{-,{3 over 10}}}^{-1/3} - 1 - {1 over 10}}
              =
              bbx{10pars{10 over 7}^{1/3} - 11}
              \[5mm] approx & 0.2625
              end{align}






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                newcommand{dd}{mathrm{d}}
                newcommand{ds}[1]{displaystyle{#1}}
                newcommand{expo}[1]{,mathrm{e}^{#1},}
                newcommand{ic}{mathrm{i}}
                newcommand{mc}[1]{mathcal{#1}}
                newcommand{mrm}[1]{mathrm{#1}}
                newcommand{pars}[1]{left(,{#1},right)}
                newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                newcommand{verts}[1]{leftvert,{#1},rightvert}$
                begin{align}
                &bbox[10px,#ffd]{sum_{n = 1}^{infty}{prod_{k = 1}^{n}pars{3k + 1} over 10^{n}pars{n + 1}!}} =
                sum_{n = 2}^{infty}{3^{n - 1}
                prod_{k = 1}^{n - 1}pars{k + 1/3} over 10^{n - 1},n!}
                \[5mm] = &
                {10 over 3}sum_{n = 2}^{infty}pars{3 over 10}^{n},
                {Gammapars{4/3 + bracks{n - 1}}/Gammapars{4/3} over n!}
                \[5mm] = &
                {10 over 3},{pars{-2/3}! over Gammapars{4/3}}
                sum_{n = 2}^{infty}pars{3 over 10}^{n},
                {pars{n - 2/3}! over n!pars{-2/3}!}
                \[5mm] = &
                {10 over 3},{Gammapars{1/3} over pars{1/3}Gammapars{1/3}}
                sum_{n = 2}^{infty}pars{3 over 10}^{n},{n - 2/3 choose n}
                \[5mm] = &
                10sum_{n = 2}^{infty}pars{3 over 10}^{n}
                bracks{{-1/3 choose n}pars{-1}^{n}}
                \[5mm] = &
                10bracks{%
                sum_{n = 0}^{infty}{-1/3 choose n}pars{-,{3 over 10}}^{n}
                - overbrace{-1/3 choose 0}^{ds{= 1}} -
                overbrace{-1/3 choose 1}^{ds{= -,{1 over 3}}}
                pars{-,{3 over 10}}}
                \[5mm] = &
                10braces{bracks{1 + pars{-,{3 over 10}}}^{-1/3} - 1 - {1 over 10}}
                =
                bbx{10pars{10 over 7}^{1/3} - 11}
                \[5mm] approx & 0.2625
                end{align}






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                  newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                  newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                  newcommand{dd}{mathrm{d}}
                  newcommand{ds}[1]{displaystyle{#1}}
                  newcommand{expo}[1]{,mathrm{e}^{#1},}
                  newcommand{ic}{mathrm{i}}
                  newcommand{mc}[1]{mathcal{#1}}
                  newcommand{mrm}[1]{mathrm{#1}}
                  newcommand{pars}[1]{left(,{#1},right)}
                  newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                  newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                  newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                  newcommand{verts}[1]{leftvert,{#1},rightvert}$
                  begin{align}
                  &bbox[10px,#ffd]{sum_{n = 1}^{infty}{prod_{k = 1}^{n}pars{3k + 1} over 10^{n}pars{n + 1}!}} =
                  sum_{n = 2}^{infty}{3^{n - 1}
                  prod_{k = 1}^{n - 1}pars{k + 1/3} over 10^{n - 1},n!}
                  \[5mm] = &
                  {10 over 3}sum_{n = 2}^{infty}pars{3 over 10}^{n},
                  {Gammapars{4/3 + bracks{n - 1}}/Gammapars{4/3} over n!}
                  \[5mm] = &
                  {10 over 3},{pars{-2/3}! over Gammapars{4/3}}
                  sum_{n = 2}^{infty}pars{3 over 10}^{n},
                  {pars{n - 2/3}! over n!pars{-2/3}!}
                  \[5mm] = &
                  {10 over 3},{Gammapars{1/3} over pars{1/3}Gammapars{1/3}}
                  sum_{n = 2}^{infty}pars{3 over 10}^{n},{n - 2/3 choose n}
                  \[5mm] = &
                  10sum_{n = 2}^{infty}pars{3 over 10}^{n}
                  bracks{{-1/3 choose n}pars{-1}^{n}}
                  \[5mm] = &
                  10bracks{%
                  sum_{n = 0}^{infty}{-1/3 choose n}pars{-,{3 over 10}}^{n}
                  - overbrace{-1/3 choose 0}^{ds{= 1}} -
                  overbrace{-1/3 choose 1}^{ds{= -,{1 over 3}}}
                  pars{-,{3 over 10}}}
                  \[5mm] = &
                  10braces{bracks{1 + pars{-,{3 over 10}}}^{-1/3} - 1 - {1 over 10}}
                  =
                  bbx{10pars{10 over 7}^{1/3} - 11}
                  \[5mm] approx & 0.2625
                  end{align}






                  share|cite|improve this answer











                  $endgroup$



                  $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                  newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                  newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                  newcommand{dd}{mathrm{d}}
                  newcommand{ds}[1]{displaystyle{#1}}
                  newcommand{expo}[1]{,mathrm{e}^{#1},}
                  newcommand{ic}{mathrm{i}}
                  newcommand{mc}[1]{mathcal{#1}}
                  newcommand{mrm}[1]{mathrm{#1}}
                  newcommand{pars}[1]{left(,{#1},right)}
                  newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                  newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                  newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                  newcommand{verts}[1]{leftvert,{#1},rightvert}$
                  begin{align}
                  &bbox[10px,#ffd]{sum_{n = 1}^{infty}{prod_{k = 1}^{n}pars{3k + 1} over 10^{n}pars{n + 1}!}} =
                  sum_{n = 2}^{infty}{3^{n - 1}
                  prod_{k = 1}^{n - 1}pars{k + 1/3} over 10^{n - 1},n!}
                  \[5mm] = &
                  {10 over 3}sum_{n = 2}^{infty}pars{3 over 10}^{n},
                  {Gammapars{4/3 + bracks{n - 1}}/Gammapars{4/3} over n!}
                  \[5mm] = &
                  {10 over 3},{pars{-2/3}! over Gammapars{4/3}}
                  sum_{n = 2}^{infty}pars{3 over 10}^{n},
                  {pars{n - 2/3}! over n!pars{-2/3}!}
                  \[5mm] = &
                  {10 over 3},{Gammapars{1/3} over pars{1/3}Gammapars{1/3}}
                  sum_{n = 2}^{infty}pars{3 over 10}^{n},{n - 2/3 choose n}
                  \[5mm] = &
                  10sum_{n = 2}^{infty}pars{3 over 10}^{n}
                  bracks{{-1/3 choose n}pars{-1}^{n}}
                  \[5mm] = &
                  10bracks{%
                  sum_{n = 0}^{infty}{-1/3 choose n}pars{-,{3 over 10}}^{n}
                  - overbrace{-1/3 choose 0}^{ds{= 1}} -
                  overbrace{-1/3 choose 1}^{ds{= -,{1 over 3}}}
                  pars{-,{3 over 10}}}
                  \[5mm] = &
                  10braces{bracks{1 + pars{-,{3 over 10}}}^{-1/3} - 1 - {1 over 10}}
                  =
                  bbx{10pars{10 over 7}^{1/3} - 11}
                  \[5mm] approx & 0.2625
                  end{align}







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 29 '18 at 18:39

























                  answered Feb 1 '18 at 20:00









                  Felix MarinFelix Marin

                  68.7k7109146




                  68.7k7109146















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