3

$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{m²n}{n3^m +m3^n}$.

I replaced m by n,n by m and sum both which gives term $\frac{mn(m+n)}{n3^m +m3^n}$.how to do further?

Ninad Munshi
  • 34,407
user69608
  • 858
  • Hello OP, please check if your problem has any errors. There is a nice solution if the problem was written slightly differently, but until you confirm otherwise I have rolled back the edit to your original question. – Ninad Munshi Sep 08 '20 at 07:54
  • @NinadMunshi there may be printing mistake i am not sure of,but i got the problem in this form. – user69608 Sep 08 '20 at 08:00
  • @user69608 Maybe the same source gives the final/expected answer as well. Share with us if you have this. – Z Ahmed Sep 08 '20 at 09:52

1 Answers1

7

A closely related summation doable by hand is when $3^m$ multiplies the denominator of OP: $$S=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{m²n}{3^m(n3^m +m3^n)}~~~(1)$$ $$S=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{a_m(a_m+a_n)},~~ a_k=3^k/k~~~(2)$$ Interchange $m$ and $n$, then $$S=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{a_n(a_n+a_m)}~~~(3)$$ Adding (2) and (3), we get $$2S=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{a_m a_n} =\left(\sum_ {k=0}^{\infty} \frac{1}{a_k}\right)^2= \left(\sum_ {k=0}^{\infty} \frac{k}{3^k}\right)^2.$$ Next use $$\sum_{k=1}^{\infty} kx^k=\frac{x}{(1-x)^2}, {x}<1.$$ $$\implies S=\frac{1}{2} \frac{9}{16}=\frac{9}{32}$$

Z Ahmed
  • 43,235
  • 2
    Hold on, isn't it a bit convenient to change the problem to something completely different and answer that without OP responding first? – Ninad Munshi Sep 08 '20 at 07:48
  • Otherwise it is not doable by hand. Please check. – Z Ahmed Sep 08 '20 at 07:48
  • 2
    $-1$ That's not the point. We can't just change problems to things we can do, unless you have a strong reason to think this is the problem OP meant other than "it is not doable otherwise" – Ninad Munshi Sep 08 '20 at 07:51
  • 2
    I have rolled back the edits. Please do not change OP's post to reflect the question you wanted to answer unless OP says otherwise. – Ninad Munshi Sep 08 '20 at 07:55
  • 1
    I don't see any proof that the original problem is not doable ;) – ypercubeᵀᴹ Sep 08 '20 at 08:05
  • I would love to see a solution for the case when $3^m$ does not multiply the denominator. Till the one may enjoy the doable form. – Z Ahmed Sep 08 '20 at 08:09