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?


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


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


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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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