Interchanging Summation and Integrals?

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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.