Is the closed-form of $sum_{k=1}^{infty}frac{e^{-k}}{k^2}$ known?












3












$begingroup$


I'm looking for an expression that yields the above sum.



It's a straight and simple question, feel free to move this post to any other section, if it's not in the appropriate section, I know this site can be pretty strict with questions:



Is $sum_{k=1}^{infty}frac{e^{-k}}{k^2}$ known?






sequences-and-series closed-form







share|cite|improve this question











$endgroup$








  • 12




    $begingroup$
    begin{eqnarray*} Li_2( frac{1}{e}). end{eqnarray*} en.wikipedia.org/wiki/Spence%27s_function
    $endgroup$
    – Donald Splutterwit
    Jan 1 at 23:20






  • 4




    $begingroup$
    Try evaluating $sum{x^kover k^2}$ ay $x={1over e}$
    $endgroup$
    – saulspatz
    Jan 1 at 23:21










  • $begingroup$
    If you don't like dilogarithms, follow saulspatz advice and use differentation of power series theorem
    $endgroup$
    – Jakobian
    Jan 1 at 23:31






  • 1




    $begingroup$
    WolframAlpha does not produce something helpful therefore I have doubts that a closed-form is known.
    $endgroup$
    – mrtaurho
    Jan 2 at 0:07
















3












$begingroup$


I'm looking for an expression that yields the above sum.



It's a straight and simple question, feel free to move this post to any other section, if it's not in the appropriate section, I know this site can be pretty strict with questions:



Is $sum_{k=1}^{infty}frac{e^{-k}}{k^2}$ known?






sequences-and-series closed-form







share|cite|improve this question











$endgroup$








  • 12




    $begingroup$
    begin{eqnarray*} Li_2( frac{1}{e}). end{eqnarray*} en.wikipedia.org/wiki/Spence%27s_function
    $endgroup$
    – Donald Splutterwit
    Jan 1 at 23:20






  • 4




    $begingroup$
    Try evaluating $sum{x^kover k^2}$ ay $x={1over e}$
    $endgroup$
    – saulspatz
    Jan 1 at 23:21










  • $begingroup$
    If you don't like dilogarithms, follow saulspatz advice and use differentation of power series theorem
    $endgroup$
    – Jakobian
    Jan 1 at 23:31






  • 1




    $begingroup$
    WolframAlpha does not produce something helpful therefore I have doubts that a closed-form is known.
    $endgroup$
    – mrtaurho
    Jan 2 at 0:07














3












3








3


2



$begingroup$


I'm looking for an expression that yields the above sum.



It's a straight and simple question, feel free to move this post to any other section, if it's not in the appropriate section, I know this site can be pretty strict with questions:



Is $sum_{k=1}^{infty}frac{e^{-k}}{k^2}$ known?






sequences-and-series closed-form







share|cite|improve this question











$endgroup$




I'm looking for an expression that yields the above sum.



It's a straight and simple question, feel free to move this post to any other section, if it's not in the appropriate section, I know this site can be pretty strict with questions:



Is $sum_{k=1}^{infty}frac{e^{-k}}{k^2}$ known?






sequences-and-series closed-form




sequences-and-series closed-form






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 23:22









Cheerful Parsnip

21k23498




21k23498










asked Jan 1 at 23:17









JR S.JR S.

162




162








  • 12




    $begingroup$
    begin{eqnarray*} Li_2( frac{1}{e}). end{eqnarray*} en.wikipedia.org/wiki/Spence%27s_function
    $endgroup$
    – Donald Splutterwit
    Jan 1 at 23:20






  • 4




    $begingroup$
    Try evaluating $sum{x^kover k^2}$ ay $x={1over e}$
    $endgroup$
    – saulspatz
    Jan 1 at 23:21










  • $begingroup$
    If you don't like dilogarithms, follow saulspatz advice and use differentation of power series theorem
    $endgroup$
    – Jakobian
    Jan 1 at 23:31






  • 1




    $begingroup$
    WolframAlpha does not produce something helpful therefore I have doubts that a closed-form is known.
    $endgroup$
    – mrtaurho
    Jan 2 at 0:07














  • 12




    $begingroup$
    begin{eqnarray*} Li_2( frac{1}{e}). end{eqnarray*} en.wikipedia.org/wiki/Spence%27s_function
    $endgroup$
    – Donald Splutterwit
    Jan 1 at 23:20






  • 4




    $begingroup$
    Try evaluating $sum{x^kover k^2}$ ay $x={1over e}$
    $endgroup$
    – saulspatz
    Jan 1 at 23:21










  • $begingroup$
    If you don't like dilogarithms, follow saulspatz advice and use differentation of power series theorem
    $endgroup$
    – Jakobian
    Jan 1 at 23:31






  • 1




    $begingroup$
    WolframAlpha does not produce something helpful therefore I have doubts that a closed-form is known.
    $endgroup$
    – mrtaurho
    Jan 2 at 0:07








12




12




$begingroup$
begin{eqnarray*} Li_2( frac{1}{e}). end{eqnarray*} en.wikipedia.org/wiki/Spence%27s_function
$endgroup$
– Donald Splutterwit
Jan 1 at 23:20




$begingroup$
begin{eqnarray*} Li_2( frac{1}{e}). end{eqnarray*} en.wikipedia.org/wiki/Spence%27s_function
$endgroup$
– Donald Splutterwit
Jan 1 at 23:20




4




4




$begingroup$
Try evaluating $sum{x^kover k^2}$ ay $x={1over e}$
$endgroup$
– saulspatz
Jan 1 at 23:21




$begingroup$
Try evaluating $sum{x^kover k^2}$ ay $x={1over e}$
$endgroup$
– saulspatz
Jan 1 at 23:21












$begingroup$
If you don't like dilogarithms, follow saulspatz advice and use differentation of power series theorem
$endgroup$
– Jakobian
Jan 1 at 23:31




$begingroup$
If you don't like dilogarithms, follow saulspatz advice and use differentation of power series theorem
$endgroup$
– Jakobian
Jan 1 at 23:31




1




1




$begingroup$
WolframAlpha does not produce something helpful therefore I have doubts that a closed-form is known.
$endgroup$
– mrtaurho
Jan 2 at 0:07




$begingroup$
WolframAlpha does not produce something helpful therefore I have doubts that a closed-form is known.
$endgroup$
– mrtaurho
Jan 2 at 0:07










1 Answer
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$begingroup$

Making the problem more general
$$sum_{k=1}^{infty}frac{e^{-k x}}{k^n}=text{Li}_nleft(e^{-x}right)$$where appears the polylogarithm function which is a special function.






share|cite|improve this answer









$endgroup$













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    1 Answer
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    1 Answer
    1






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    oldest

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    oldest

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    active

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    $begingroup$

    Making the problem more general
    $$sum_{k=1}^{infty}frac{e^{-k x}}{k^n}=text{Li}_nleft(e^{-x}right)$$where appears the polylogarithm function which is a special function.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Making the problem more general
      $$sum_{k=1}^{infty}frac{e^{-k x}}{k^n}=text{Li}_nleft(e^{-x}right)$$where appears the polylogarithm function which is a special function.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Making the problem more general
        $$sum_{k=1}^{infty}frac{e^{-k x}}{k^n}=text{Li}_nleft(e^{-x}right)$$where appears the polylogarithm function which is a special function.






        share|cite|improve this answer









        $endgroup$



        Making the problem more general
        $$sum_{k=1}^{infty}frac{e^{-k x}}{k^n}=text{Li}_nleft(e^{-x}right)$$where appears the polylogarithm function which is a special function.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 3:54









        Claude LeiboviciClaude Leibovici

        120k1157132




        120k1157132






























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