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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!

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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

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  • $\begingroup$ But unfortunately, the series inside the integral sign does not convergence uniformly on $\left[0,1\right]$ (or $\left[0,1\right)$ ), since it is divergent at $x=1$. However it is u.c. on any closed intervals inside $\left[0,1\right)$. $\endgroup$ – Antimonius Dec 12 '17 at 10:36

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