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Question: Calculate

$$\sum_{n\geq1}\sum_{m\geq1}\frac{mn}{n\cdot3^m+m\cdot3^n}\left(\frac{n}{3^m}+\frac{m}{3^n}\right)$$

My attempt: I let $a_k=\frac{3^k}{k}$ and wrote the sum like this

$$\frac{n}{ma_m(a_m+a_n)}+\frac{m}{na_n(a_m+a_n)}$$

Now, since the second expression is the same as the first but with $m,n$ exchanged, so the sum will be two times the sum of the first but I can't compute the first sum.

Any help on how to proceed would be appreciated.

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  • $\begingroup$ What about to change summations? $\sum_{n \ge 1} \sum_{m \le n}$ $\endgroup$
    – openspace
    Sep 11 '20 at 11:06
  • $\begingroup$ @openspace I don't get what you are implying. Aren't both of these sums independent with both $m,n$ separately going from $1$ to $\infty$? $\endgroup$
    – user35508
    Sep 11 '20 at 11:15
  • $\begingroup$ Consider plane with $m,n$ axis. Then you're trying to compute all $a(n,m)$ over this plane. F.o.a. you can see that $a(n,m) = a(m,n)$. So you just need to compute over all $m \le n$ and $n \ge 1$. And that's equal to what I've mentioned. $\endgroup$
    – openspace
    Sep 11 '20 at 12:00
  • $\begingroup$ @openspace Ok I now get what you said. But I still can't figure out how to finish $\endgroup$
    – user35508
    Sep 11 '20 at 12:05
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Maybe it will be easier.

$$\sum_{n \ge 1} \sum_{m \ge 1} a(n,m) = 2\sum_{n \ge 1} \sum_{1 \le m < n} a(n,m) + \sum_{n \ge 1} a(n,n)$$

Now we have $$\sum_{n \ge 1} a(n,n) = \sum_{n \ge 1} \dfrac{n}{3^{2n}} $$

and for first part:

$$\sum_{n \ge 1} \sum_{1 \le m < n} \dfrac{mn}{n 3^m + m3^n}\left(\dfrac{n}{3^m} +\dfrac{m}{3^n}\right)$$

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  • $\begingroup$ This is more appropriate as a comment. $\endgroup$
    – Integrand
    Sep 11 '20 at 13:15

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