# Interchanging Summation and Integral

Lets Think of this: I posted a question and got a wonderful answer by a smart user, but I couldn’t understand part of the method.

\begin{align} \sum_{n=1}^{\infty}\frac{1}{9n^2 + 3n – 2} &=\frac{1}{3}\sum_{n=1}^{\infty}\left(\frac{1}{3n – 1}-\frac{1}{3n + 2}\right)\\\\ &=\frac{1}{3}\sum_{n=1}^{\infty}\int_0^1\left(x^{3n-2}-x^{3n+1}\right){\rm d}x\\\\ &=\frac{1}{3}\int_0^1\sum_{n=1}^{\infty}\left(x^{3n-2}-x^{3n+1}\right){\rm d}x\\\\\end{align}

How is it possible to interchange the summation and integral?

Thanks!

And in general, for what does this apply? Thank you very much!

#### Solutions Collecting From Web of "Interchanging Summation and Integral"

You’re allowed to do this anytime the series is uniformly convergent.

In uniform convergence, you tell me the $\epsilon$ that you want the entire partial sum to be within for the entire graph, and I give you an $M$ that guarantees you can get within $\epsilon$ if you choose $n > M$, but the $M$ must work anywhere on the graph, and not be dependent on which point you choose. In regular (called pointwise) convergence, the $\epsilon-M$ guarantee can use a different $M$ at different points on the graph.

Pointwise vs. Uniform Convergence

https://proofwiki.org/wiki/Definition:Uniform_Convergence