How to find the sum of another series












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How do I find the sum of a series



$$sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}$$






sequences-and-series







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  • $begingroup$
    Try something along these lines.
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    – Clement C.
    Dec 24 '18 at 20:52






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    Dec 25 '18 at 15:24
















0












$begingroup$


How do I find the sum of a series



$$sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}$$






sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    Try something along these lines.
    $endgroup$
    – Clement C.
    Dec 24 '18 at 20:52






  • 1




    $begingroup$
    Who DELETED math.stackexchange.com/questions/3050493??
    $endgroup$
    – cgiovanardi
    Dec 25 '18 at 15:24














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How do I find the sum of a series



$$sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}$$






sequences-and-series







share|cite|improve this question









$endgroup$




How do I find the sum of a series



$$sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}$$






sequences-and-series




sequences-and-series






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asked Dec 24 '18 at 20:49









Some studentSome student

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  • $begingroup$
    Try something along these lines.
    $endgroup$
    – Clement C.
    Dec 24 '18 at 20:52






  • 1




    $begingroup$
    Who DELETED math.stackexchange.com/questions/3050493??
    $endgroup$
    – cgiovanardi
    Dec 25 '18 at 15:24


















  • $begingroup$
    Try something along these lines.
    $endgroup$
    – Clement C.
    Dec 24 '18 at 20:52






  • 1




    $begingroup$
    Who DELETED math.stackexchange.com/questions/3050493??
    $endgroup$
    – cgiovanardi
    Dec 25 '18 at 15:24
















$begingroup$
Try something along these lines.
$endgroup$
– Clement C.
Dec 24 '18 at 20:52




$begingroup$
Try something along these lines.
$endgroup$
– Clement C.
Dec 24 '18 at 20:52




1




1




$begingroup$
Who DELETED math.stackexchange.com/questions/3050493??
$endgroup$
– cgiovanardi
Dec 25 '18 at 15:24




$begingroup$
Who DELETED math.stackexchange.com/questions/3050493??
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– cgiovanardi
Dec 25 '18 at 15:24










2 Answers
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Write $i=x,,j=y-x,,k=z-y$ so your sum is $sum_{ijk}frac{1}{60^i 20^j 5^k}$, all indices starting at $1$. But $sum_ifrac{1}{n^i}=frac{1}{n-1}$, so the result is $frac{1}{59times 19times 4}$.






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    0












    $begingroup$

    Direct calculation through expansion:
    $$begin{align}sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}=
    &color{red}{frac13left[frac1{4^2}left(frac1{5^3}+frac1{5^4}+cdotsright)+frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+cdotsright]}+\
    +&color{blue}{frac1{3^2}left[frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+frac1{4^4}left(frac1{5^5}+frac1{5^6}+cdotsright)+cdotsright]}+cdots=\
    =&color{red}{frac13left[frac1{4^2}cdot frac{1}{4cdot 5^2}+frac1{4^3}cdot frac1{4cdot 5^3}+cdotsright]}+\
    +&color{blue}{frac1{3^2}left[frac1{4^3}cdot frac1{4cdot 5^3}+frac1{4^4}cdot frac1{4cdot 5^4}+cdotsright]}+cdots=\
    =&color{red}{frac13cdot frac{1}{4^3cdot 5^2}cdot frac{4cdot 5}{19}}+\
    +&color{blue}{frac1{3^2}cdot frac{1}{4^4cdot 5^3}cdot frac{4cdot 5}{19}}+cdots=\
    =&frac{1}{3cdot 4^2cdot 5cdot 19}cdot frac{3cdot 4cdot 5}{59}=\
    =&frac1{4cdot 19cdot 59}.end{align}$$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






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      active

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      4












      $begingroup$

      Write $i=x,,j=y-x,,k=z-y$ so your sum is $sum_{ijk}frac{1}{60^i 20^j 5^k}$, all indices starting at $1$. But $sum_ifrac{1}{n^i}=frac{1}{n-1}$, so the result is $frac{1}{59times 19times 4}$.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Write $i=x,,j=y-x,,k=z-y$ so your sum is $sum_{ijk}frac{1}{60^i 20^j 5^k}$, all indices starting at $1$. But $sum_ifrac{1}{n^i}=frac{1}{n-1}$, so the result is $frac{1}{59times 19times 4}$.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Write $i=x,,j=y-x,,k=z-y$ so your sum is $sum_{ijk}frac{1}{60^i 20^j 5^k}$, all indices starting at $1$. But $sum_ifrac{1}{n^i}=frac{1}{n-1}$, so the result is $frac{1}{59times 19times 4}$.






          share|cite|improve this answer









          $endgroup$



          Write $i=x,,j=y-x,,k=z-y$ so your sum is $sum_{ijk}frac{1}{60^i 20^j 5^k}$, all indices starting at $1$. But $sum_ifrac{1}{n^i}=frac{1}{n-1}$, so the result is $frac{1}{59times 19times 4}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 24 '18 at 20:54









          J.G.J.G.

          34.4k23252




          34.4k23252























              0












              $begingroup$

              Direct calculation through expansion:
              $$begin{align}sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}=
              &color{red}{frac13left[frac1{4^2}left(frac1{5^3}+frac1{5^4}+cdotsright)+frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+cdotsright]}+\
              +&color{blue}{frac1{3^2}left[frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+frac1{4^4}left(frac1{5^5}+frac1{5^6}+cdotsright)+cdotsright]}+cdots=\
              =&color{red}{frac13left[frac1{4^2}cdot frac{1}{4cdot 5^2}+frac1{4^3}cdot frac1{4cdot 5^3}+cdotsright]}+\
              +&color{blue}{frac1{3^2}left[frac1{4^3}cdot frac1{4cdot 5^3}+frac1{4^4}cdot frac1{4cdot 5^4}+cdotsright]}+cdots=\
              =&color{red}{frac13cdot frac{1}{4^3cdot 5^2}cdot frac{4cdot 5}{19}}+\
              +&color{blue}{frac1{3^2}cdot frac{1}{4^4cdot 5^3}cdot frac{4cdot 5}{19}}+cdots=\
              =&frac{1}{3cdot 4^2cdot 5cdot 19}cdot frac{3cdot 4cdot 5}{59}=\
              =&frac1{4cdot 19cdot 59}.end{align}$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Direct calculation through expansion:
                $$begin{align}sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}=
                &color{red}{frac13left[frac1{4^2}left(frac1{5^3}+frac1{5^4}+cdotsright)+frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+cdotsright]}+\
                +&color{blue}{frac1{3^2}left[frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+frac1{4^4}left(frac1{5^5}+frac1{5^6}+cdotsright)+cdotsright]}+cdots=\
                =&color{red}{frac13left[frac1{4^2}cdot frac{1}{4cdot 5^2}+frac1{4^3}cdot frac1{4cdot 5^3}+cdotsright]}+\
                +&color{blue}{frac1{3^2}left[frac1{4^3}cdot frac1{4cdot 5^3}+frac1{4^4}cdot frac1{4cdot 5^4}+cdotsright]}+cdots=\
                =&color{red}{frac13cdot frac{1}{4^3cdot 5^2}cdot frac{4cdot 5}{19}}+\
                +&color{blue}{frac1{3^2}cdot frac{1}{4^4cdot 5^3}cdot frac{4cdot 5}{19}}+cdots=\
                =&frac{1}{3cdot 4^2cdot 5cdot 19}cdot frac{3cdot 4cdot 5}{59}=\
                =&frac1{4cdot 19cdot 59}.end{align}$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Direct calculation through expansion:
                  $$begin{align}sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}=
                  &color{red}{frac13left[frac1{4^2}left(frac1{5^3}+frac1{5^4}+cdotsright)+frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+cdotsright]}+\
                  +&color{blue}{frac1{3^2}left[frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+frac1{4^4}left(frac1{5^5}+frac1{5^6}+cdotsright)+cdotsright]}+cdots=\
                  =&color{red}{frac13left[frac1{4^2}cdot frac{1}{4cdot 5^2}+frac1{4^3}cdot frac1{4cdot 5^3}+cdotsright]}+\
                  +&color{blue}{frac1{3^2}left[frac1{4^3}cdot frac1{4cdot 5^3}+frac1{4^4}cdot frac1{4cdot 5^4}+cdotsright]}+cdots=\
                  =&color{red}{frac13cdot frac{1}{4^3cdot 5^2}cdot frac{4cdot 5}{19}}+\
                  +&color{blue}{frac1{3^2}cdot frac{1}{4^4cdot 5^3}cdot frac{4cdot 5}{19}}+cdots=\
                  =&frac{1}{3cdot 4^2cdot 5cdot 19}cdot frac{3cdot 4cdot 5}{59}=\
                  =&frac1{4cdot 19cdot 59}.end{align}$$






                  share|cite|improve this answer









                  $endgroup$



                  Direct calculation through expansion:
                  $$begin{align}sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}=
                  &color{red}{frac13left[frac1{4^2}left(frac1{5^3}+frac1{5^4}+cdotsright)+frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+cdotsright]}+\
                  +&color{blue}{frac1{3^2}left[frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+frac1{4^4}left(frac1{5^5}+frac1{5^6}+cdotsright)+cdotsright]}+cdots=\
                  =&color{red}{frac13left[frac1{4^2}cdot frac{1}{4cdot 5^2}+frac1{4^3}cdot frac1{4cdot 5^3}+cdotsright]}+\
                  +&color{blue}{frac1{3^2}left[frac1{4^3}cdot frac1{4cdot 5^3}+frac1{4^4}cdot frac1{4cdot 5^4}+cdotsright]}+cdots=\
                  =&color{red}{frac13cdot frac{1}{4^3cdot 5^2}cdot frac{4cdot 5}{19}}+\
                  +&color{blue}{frac1{3^2}cdot frac{1}{4^4cdot 5^3}cdot frac{4cdot 5}{19}}+cdots=\
                  =&frac{1}{3cdot 4^2cdot 5cdot 19}cdot frac{3cdot 4cdot 5}{59}=\
                  =&frac1{4cdot 19cdot 59}.end{align}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 25 '18 at 17:40









                  farruhotafarruhota

                  22.5k2942




                  22.5k2942






























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