Is $\sum_{j=1}^\infty\sum_{n=1}^\infty\left(\frac{e^{-j/n}}{n^2}-\frac{e^{-n/j}}{j^2}\right)=\gamma ?$ A friend proposed the following problem:
$$\sum_{j=1}^\infty\sum_{n=1}^\infty\left(\frac{e^{-j/n}}{n^2}-\frac{e^{-n/j}}{j^2}\right)=\gamma,$$
where $\gamma$ is the Euler-Mascheroni constant.
The result I got is zero and here is what I did:
$$\sum_{j=1}^\infty\sum_{n=1}^\infty\left(\frac{e^{-j/n}}{n^2}-\frac{e^{-n/j}}{j^2}\right)=\sum_{j=1}^\infty\sum_{n=1}^\infty\frac{e^{-j/n}}{n^2}-\sum_{j=1}^\infty\sum_{n=1}^\infty\frac{e^{-n/j}}{j^2}$$
(swap $n$ and $j $ in the first double sum then change the order of summations)
$$=\sum_{j=1}^\infty\sum_{n=1}^\infty\frac{e^{-n/j}}{j^2}-\sum_{j=1}^\infty\sum_{n=1}^\infty\frac{e^{-n/j}}{j^2}=0.$$
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.

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
