A friend proposed the following problem:


where $\gamma$ is the Euler-Mascheroni constant.

The result I got is zero and here is what I did:


(swap $n$ and $j $ in the first double sum then change the order of summations)


Even Mathematica gives zero.

Is it possible that this double sum equals $\gamma$ or my friend could be wrong?

Thanks to @Thomas Andrews for noticing that its not valid to break up the summand as the double sums don't converge and that tells us that we can't always rely on results given by Mathematica or Wolfram.

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    $\begingroup$ As a rule $$\sum (a_{ij}-b_{ij})=\sum a_{ij}-\sum b_{ij}$$ only if the sums in the right converge. Do we know that here? $\endgroup$ Jan 8 at 20:13
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    $\begingroup$ Here. $$\sum_n e^{-n/j}/j^2=\frac{1}{j^2(e^{1/j}-1)}\sim \frac{1}{j}$$ so the double sums don't converge. $\endgroup$ Jan 8 at 20:17
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    $\begingroup$ (1) Do not put meta commentary in your question. (This is the reason ArcticChar edited your question). (2) Large amounts of bold are distracting, and the remainder of my edit was to improve the grammar and writing style of your question. These edits, in my opinion, improve your question. $\endgroup$
    – Xander Henderson
    Jan 18 at 17:59
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    $\begingroup$ @AliShadhar The paragraph which ArcticChar deleted contained zero mathematical content. As such, it did not belong in your question, and it was removed. I cannot possibly read ArcticChar's mind, hence I do not know if they deleted your commentary "because he felt that applies to him". I have no idea what motived them. Their motivation is irrelevant. Your meta commentary was inappropriate, and was appropriately removed. $\endgroup$
    – Xander Henderson
    Jan 18 at 18:10
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    $\begingroup$ Text within a \text{} block does not linewrap. Thus your edit creates text which is not easily read on narrow displays, or by people who are using large fonts. Putting that text in a \text{} block actively disimproves the readability and accessibility of your question. $\endgroup$
    – Xander Henderson
    Jan 22 at 12:54

2 Answers 2


Here is a relatively elementary approach. $$ \begin{align} \lim_{n\to\infty}\sum_{j=1}^{n-1}\sum_{k=n}^\infty\frac{e^{-j/k}}{k^2} &=\lim_{n\to\infty}\sum_{j=1}^{n-1}\int_n^\infty\frac{e^{-j/x}}{x^2}\,\mathrm{d}x\tag{1a}\\ &=\lim_{n\to\infty}\sum_{j=1}^{n-1}\int_0^{1/n}e^{-jx}\,\mathrm{d}x\tag{1b}\\ &=\lim_{n\to\infty}\sum_{j=1}^{n-1}\frac nj\left(1-e^{-j/n}\right)\frac1n\tag{1c}\\ &=\int_0^1\frac{1-e^{-x}}{x}\,\mathrm{d}x\tag{1d}\\[3pt] &=-\int_0^1\log(x)\,e^{-x}\,\mathrm{d}x\tag{1e} \end{align} $$ Explanation:
$\text{(1a)}$: approximate the sum with an integral:
$\phantom{\text{(1a):}}$ since $x\ge j$, the summand is decreasing and less than $\frac1{n^2}$
$\phantom{\text{(1a):}}$ the step size is $1$, so the error in the approximation is less than $\frac1{n^2}\cdot1$
$\phantom{\text{(1a):}}$ the error in summing $n-1$ of these is less than $\frac1n$
$\text{(1b)}$: substitute $x\mapsto\frac1x$
$\text{(1c)}$: integrate
$\text{(1d)}$: Riemann sum
$\text{(1e)}$: integrate by parts $$ \begin{align} \lim_{n\to\infty}\sum_{j=1}^{n-1}\sum_{k=n}^\infty\frac{e^{-k/j}}{j^2} &=\lim_{n\to\infty}\sum_{j=1}^{n-1}\frac{e^{-n/j}}{j^2\left(1-e^{-1/j}\right)}\tag{2a}\\ &=\lim_{n\to\infty}\sum_{j=1}^{n-1}\frac{e^{-n/j}}{j/n}\frac1n\tag{2b}\\ &=\int_0^1\frac{e^{-1/x}}{x}\,\mathrm{d}x\tag{2c}\\[3pt] &=\int_1^\infty\frac{e^{-x}}{x}\,\mathrm{d}x\tag{2d}\\[3pt] &=\int_1^\infty\log(x)\,e^{-x}\,\mathrm{d}x\tag{2e} \end{align} $$ Explanation:
$\text{(2a)}$: sum the geometric series
$\text{(2b)}$: since $j-\frac12\le j^2\left(1-e^{-1/j}\right)\le j$, the sum of the errors is less than
$\phantom{\text{(2b):}}$ $\sum\limits_{j=1}^{n-1}\frac{e^{-n/j}}{j(2j-1)}\le\frac1{n^2}\sum\limits_{j=1}^{n-1}\frac{e^{-n/j}}{(j/n)^2}\le\frac4{e^2n}$ because $\frac{e^{-n/j}}{(j/n)^2}=u^2e^{-u}\le\frac4{e^2}$
$\text{(2c)}$: Riemann sum
$\text{(2d)}$: substitute $x\mapsto\frac1x$
$\text{(2e)}$: integrate by parts $$ \begin{align} \sum_{j=1}^\infty\sum_{k=1}^\infty\left(\frac{e^{-j/k}}{k^2}-\frac{e^{-k/j}}{j^2}\right) &=\lim_{n\to\infty}\sum_{j=1}^{n-1}\sum_{k=1}^\infty\left(\frac{e^{-j/k}}{k^2}-\frac{e^{-k/j}}{j^2}\right)\tag{3a}\\ &=\lim_{n\to\infty}\sum_{j=1}^{n-1}\sum_{k=n}^\infty\left(\frac{e^{-j/k}}{k^2}-\frac{e^{-k/j}}{j^2}\right)\tag{3b}\\ &=-\int_0^\infty\log(x)\,e^{-x}\,\mathrm{d}x\tag{3c}\\[9pt] &=\gamma\tag{3d} \end{align} $$ Explanation:
$\text{(3a)}$: write infinite sum as a limit
$\text{(3b)}$: subtract $\sum\limits_{j=1}^{n-1}\sum\limits_{k=1}^{n-1}\left(\frac{e^{-j/k}}{k^2}-\frac{e^{-k/j}}{j^2}\right)=0$
$\text{(3c)}$: apply $(1)$ and $(2)$
$\text{(3d)}$: apply this answer


Here's the solution I have. Take $1<c<2$ and consider the integral $$I=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(s)\zeta(s)\zeta(2-s)\,ds.$$

Let's compute it in two ways.

First, $I$ is the sum of the residues of the integrand at all integer $s\leqslant1$ (consider integration along the boundary of $[-N-1/2,c]+i[-R,R]$ with $R>0$ and $N$ a positive integer, and take $R,N\to\infty$; the integral along the "right side of the rectangle" tends to $I$, and the remaining integrals vanish). The residue at $s=1$ (a double pole) may be computed using \begin{align*} \Gamma(s)&=1-\gamma(s-1)+o(s-1),\\ \zeta(s)&=(s-1)^{-1}+\gamma+o(1),\\ \zeta(2-s)&=-(s-1)^{-1}+\gamma+o(1), \end{align*} hence $\Gamma(s)\zeta(s)\zeta(2-s)=(s-1)^{-2}\big({-1}+\gamma(s-1)+o(s-1)\big)$ and the residue equals $\gamma$. And, for a nonnegative integer $n$, we have $$\newcommand{\res}{\operatorname*{Res}}\res_{s=-n}\Gamma(s)\zeta(s)\zeta(2-s)=\frac{(-1)^n}{n!}\zeta(-n)\zeta(n+2)=\frac{B_{n+1}}{(n+1)!}\zeta(n+2)$$ so that, using the generating function of the Bernoulli numbers, \begin{align*} I&=\gamma+\sum_{n=0}^\infty\frac{B_{n+1}}{(n+1)!}\zeta(n+2)=\gamma+\sum_{n=1}^\infty\frac{B_n}{n!}\sum_{j=1}^\infty\frac1{j^{n+1}} \\&=\gamma+\sum_{j=1}^\infty\frac1j\left(\frac{1/j}{e^{1/j}-1}-1\right)=\gamma+\sum_{j=1}^\infty\left(-\frac1j+\frac1{j^2}\sum_{n=1}^\infty e^{-n/j}\right). \end{align*}

Second (the implied uniform convergence is not hard to see; recall $1<c<2$), $$I=\sum_{j=1}^\infty I_j,\quad I_j:=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(s)j^{-s}\zeta(2-s)\,ds.$$ Now $I_j$ may be computed the same way. The residue of the integrand at $s=1$ equals $-1/j$, and the residue at $s=-k$ (for a nonnegative integer $k$) is $\zeta(k+2)(-j)^k/k!$, so that $$I=\sum_{j=1}^\infty\left(-\frac1j+\sum_{k=0}^\infty\frac{(-j)^k}{k!}\sum_{n=1}^\infty\frac1{n^{k+2}}\right)=\sum_{j=1}^\infty\left(-\frac1j+\sum_{n=1}^\infty\frac{e^{-j/n}}{n^2}\right).$$

Comparing these two expressions for $I$, we get the expected result.


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