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The following identity seems to hold for $a>1$ :

$$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{ \frac{a^m}{m}\left( \frac{a^m}{m}+ \frac{a^n}{n} \right) } = \frac{a^2}{2(a-1)^4}$$

I've tested this with Maple for various values of $a$, but I haven't been able to make any headway on proving it. My usual tricks don't seem to work. I'd be grateful for any ideas or for a proof.

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    $\begingroup$ Multiply both sides by $2$ and replace one of the (two) double sums with the same double sum except with $n$ and $m$ changed. You then wish to show $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{mn}{a^{mn}} = \frac{a^2}{(a-1)^4}.$$ $\endgroup$
    – mathworker21
    Commented Nov 12, 2022 at 0:36
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    $\begingroup$ @mathworker21 Your $a^{mn}$ in the denominator should be $a^{m+n}$ I think, and then your sum is just $ \sum_{n=1}^{\infty} n/a^n$, squared. This sum is $a/(a-1)^2$ by, say, differentiating $\sum_{n=1}^{\infty} 1/a^n= 1/(a-1)$. $\endgroup$
    – Ofir Gorodetsky
    Commented Nov 12, 2022 at 1:46
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    $\begingroup$ @mathworker21, why don't you post your comment as an answer? I'll upvote it. $\endgroup$
    – Daniele Tampieri
    Commented Nov 12, 2022 at 8:55
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    $\begingroup$ Yes, @mathworker, I think you should write this as an answer and I will accept it. Great job. So easy when you lay it out! I'm a little embarrassed, but I'm still delighted by your answer. $\endgroup$
    – Jim Bryan
    Commented Nov 12, 2022 at 21:59

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