Is the closed-form of $sum_{k=1}^{infty}frac{e^{-k}}{k^2}$ known?
$begingroup$
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
edited Jan 1 at 23:22
Cheerful Parsnip
21k23498
asked Jan 1 at 23:17
JR S.JR S.
162
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add a comment |
$begingroup$
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
edited Jan 1 at 23:22
Cheerful Parsnip
21k23498
asked Jan 1 at 23:17
JR S.JR S.
162
$endgroup$
12
$begingroup$
begin{eqnarray*} Li_2( frac{1}{e}). end{eqnarray*} en.wikipedia.org/wiki/Spence%27s_function
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– Donald Splutterwit
Jan 1 at 23:20
4
$begingroup$
Try evaluating $sum{x^kover k^2}$ ay $x={1over e}$
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– 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
add a comment |
$begingroup$
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
edited Jan 1 at 23:22
Cheerful Parsnip
21k23498
asked Jan 1 at 23:17
JR S.JR S.
162
$endgroup$
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
edited Jan 1 at 23:22
Cheerful Parsnip
21k23498
asked Jan 1 at 23:17
JR S.JR S.
162
edited Jan 1 at 23:22
Cheerful Parsnip
21k23498
asked Jan 1 at 23:17
JR S.JR S.
162
edited Jan 1 at 23:22
Cheerful Parsnip
21k23498
edited Jan 1 at 23:22
Cheerful Parsnip
21k23498
edited Jan 1 at 23:22
Cheerful Parsnip
21k23498
21k23498
asked Jan 1 at 23:17
JR S.JR S.
162
asked Jan 1 at 23:17
JR S.JR S.
162
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
add a comment |
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
add a comment |
1 Answer
1
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oldest
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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.
answered Jan 2 at 3:54
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
add a comment |
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1 Answer
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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.
answered Jan 2 at 3:54
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 2 at 3:54
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 2 at 3:54
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
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.
answered Jan 2 at 3:54
Claude LeiboviciClaude Leibovici
120k1157132
answered Jan 2 at 3:54
Claude LeiboviciClaude Leibovici
120k1157132
answered Jan 2 at 3:54
Claude LeiboviciClaude Leibovici
120k1157132
answered Jan 2 at 3:54
Claude LeiboviciClaude Leibovici
120k1157132
120k1157132
add a comment |
add a comment |
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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