MHB Interchanging Summation and Integrals?

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The discussion centers on the interchange of summation and integration in a mathematical expression involving series and integrals. The user seeks clarification on the theorem or property that permits this interchange. A response provides a link to a resource detailing the relevant theorem, suggesting that further information can be found through a Google search. The conversation emphasizes the importance of understanding the criteria for such mathematical operations. This interchange is a common topic in analysis and requires familiarity with specific convergence conditions.
Amad27
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Hello,

Suppose we have:

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

How can you interchange the summation and integral?

what theorem allows this (or property)?? Thanks!
 
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Olok said:
Hello,

Suppose we have:

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

How can you interchange the summation and integral?

what theorem allows this (or property)?? Thanks!

Hi Olok, :)

Here's a link containing the theorem you are looking for.

criterion for interchanging summation and integration | planetmath.org

Also a Google search will give you a lot of places where this is discussed.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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