# Summation manipulation.

How can this be true?

\begin{align*} \sum_{n=1}^\infty \sum_{k=n}^\infty \frac 1{k^3} &= \sum_{k=1}^\infty \sum_{n=1}^{k} \frac 1{k^3}\\ \end{align*}

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On the left-hand side, imagine the following infinite sum of infinite sums:

\begin{aligned} \frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots\\ \frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots\\ \frac{1}{3^3}+\frac{1}{4^3}+\cdots\\ \frac{1}{4^3}+\cdots \end{aligned}

This is equal to the infinite sum $$\frac{1}{1^3}+\frac{2}{2^3}+\frac{3}{3^3}+\frac{4}{4^3}+\cdots$$ or $$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots$$

Now look at the right-hand side. The expression $\frac{1}{k^3}$ does not depend on $n$ at all, so the inner sum is just $\frac{k}{k^3}$, and $$\sum_{k=1}^\infty \frac{k}{k^3} = \sum_{k=1}^\infty \frac{1}{k^2}$$

which is the same as the left-hand side, as we showed above.

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$\sum_{k=1}^\infty \frac k{k^3}=\frac{\pi^2}6$ –  draks ... Sep 27 '12 at 13:44
Here's a better way to view it: the sum on the left sums the rows in your array; the one on the right sums the columns. –  Michael Hardy Sep 27 '12 at 18:09
@MichaelHardy I like that, thanks. –  Richard Sullivan Sep 27 '12 at 18:53

Let $a_{n,k}:=\frac 1{k^3}\chi_{\{(n,k),n\leq k\}}\geq 0$. The first sum is $$\sum_{n=1}^{+\infty}\sum_{k=1}^{+\infty}a_{n,k}$$ and the second one is $$\sum_{k=1}^{+\infty}\sum_{n=1}^{+\infty}a_{n,k}.$$ Switching the two sums is allowed as we deal with non-negative terms.

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