Question

The Riemann Sum m=1 to inf. of the Riemann Sum n=1 to inf. of n(m^2) / (3^m)(m3^n + n3^m)

Answer 1

\[\sum_{m=1}^\infty\sum_{n=1}^\infty \frac{nm^2}{3^m(m3^n+n3^m)}~~?\]

Answer 2

O_O the confusion has never been more real

Answer 3

I was thinking a double integral...
the wording is wierd (haven't seen such wording)

Answer 4

well, not directly double integral, but..

Answer 5

I've definitely seen this sum before... You can compute it with a neat averaging trick.

Answer 6

well, series and intyegeral don't have same values.... (do they?)
summing areas, not integers

Answer 7

so i guess sigma isn't a good interpretation... (but, I could be wrong)

Answer 8

I forget the exact details of the trick, but I think it goes like this. You swap \(m\) and \(n\) (after verifying that the two sums are the same), add them together, then halve it.
\[\begin{align*}\sum_{m=1}^\infty\sum_{n=1}^\infty \frac{nm^2}{3^m(m3^n+n3^m)}+\sum_{m=1}^\infty\sum_{n=1}^\infty \frac{n^2m}{3^n(n3^m+m3^n)}&=\sum_{m=1}^\infty\sum_{n=1}^\infty \frac{nm}{3^{m+n}}\\[2ex]
&=\left(\sum_{n=1}^\infty\frac{n}{3^n}\right)^2\end{align*}\]so the original sum should evaluate to half the value of the squared series.