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Compute the infinite sum $$\sum_{n=1}^{\infty}\sum_{k=0}^{\infty}\frac{n+k+1}{nk(n+k)^2}$$. I calculated the first part and got $$\sum_{k=1}^{\infty}\frac{n+k+1}{nk(n+k)^2}=\sum_{j=0}^{j=n}\frac{1}{n^2j}+\sum_{j=0}^{j=n}\frac{1}{n^3}{j}+\sum_{j=0}^{j=n}+\frac{1}{n^2j^2}-\frac{1}{n}\sum_{k=0}^{\infty}\frac{1}{k^2}$$. But don't know how to proceed.

user771946
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    The expression under the double sum makes no sense for $k=0$. Please adjust. Also, would you please add more context to the question - which is the source, e.g. author, year, level, country, etc. is it from a book, a test, a contest... ?! – dan_fulea Apr 08 '22 at 10:00
  • Do you mind spelling out your steps? For me it's not obvious how you got your result. And is the $+$ behind the second-to-last $Σ$ intentional? – flukx Apr 08 '22 at 10:13

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Since $$ \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{n+k+1}{nk(n+k)^2}=\underbrace{\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{nk(n+k)}}_{I}+\underbrace{\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{nk(n+k)^2}}_{J} $$ , $$ \begin{aligned} I&=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n k}\int_0^{{\infty}}e^{-(n+k)t}dt \\&=\int_0^{{\infty}} \ln^2(1-e^{-t})dt \\&=\int_0^1 \frac{\ln^2(1-x)}{x}dx=\int_0^1\frac{\ln^2 x}{1-x}dx \\&=\sum_{n=0}^{\infty}\int_0^1 x^n \ln^2 x dx=\sum_{n=0}^{\infty}\frac{2}{(n+1)^3} \\&=2\zeta(3) \end{aligned} $$ Similarly, $$ J=\int_0^1 t \ln^2(1-e^{-t})=-\int_0^1 \frac{\ln x \ln^2(1-x)}{x}dx $$ From here,we know $$ J=\frac{\pi ^4}{180} $$ Combine the results above,we get $$ \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{n+k+1}{nk(n+k)^2}=2 \zeta (3)+\frac{\pi ^4}{180} $$

Wahrscheinlichkeitsrechnung
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