Jun 29, 2012

A:

you can find all the documentation for arcobjects.net SDK here.

Q:

Working with a series with an infinite sum

I’m having trouble solving this problem
$$\sum_{i=1}^\infty \sum_{j=1}^\infty \frac1{i^2+j^2}$$
I can’t think of a way to evaluate this problem using only sigma, I know that I can use Abel summation, but I don’t know if this is the right approach.

A:

Use the Cauchy product. It works with a formal product
$$f(x)g(x)=\sum_{n\geq 0}\left(\sum_{k\geq 0}\frac{f^{(k)}(0)}{k!}\right)x^n$$

A:

Note that $i^{ -2}=\frac{i^2}{i^2}=\frac 1 i$, so
$$\sum_{i=1}^\infty \sum_{j=1}^\infty \frac1{i^2+j^2}=\sum_{i=1}^\infty\sum_{j=1}^\infty \frac{1}{i^2+j^2}-\sum_{i=1}^\infty\frac{1}{i}=\sum_{i=1}^\infty\sum_{j=1}^\infty \frac1{i^2+j^2}-\frac{\pi^2}{6}$$
Moreover, by the Cauchy product
$$\frac{\pi^2}{6}=\pi^2\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}=\sum_{i=1}^\infty \sum_{j=1}^\infty \frac1{i^2+j^2} \frac{(-1)^i}{(2i)!}$$

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