I'm trying to show that:
$\int_{0}^{\infty} e^{-u} \, \sum_{n=0}^{\infty} \dfrac{u^n}{2^{n} \, (n!)^2} \; du \;= \; \sum_{n=0}^{\infty} \dfrac{1}{2^{n}\, (n!)^2} \; \int_{0}^{\infty} e^{-u} \, u^n \, du$
I found this (in following YouTube-video: https://youtu.be/N-e5MJ3tHOk):
Using the triangle inequality we get
The last sum looks like Absolute convergence. So if $\sum_{n=0}^\infty |f_n(x)|$ converges for all necessary x values, we can us this sum as our dominating function $g(x)$ to justify the dominated convergence theorem.
So, back to my original problem, I try to use $\sum_{n=0}^\infty |\dfrac{u^n}{2^{n} \, (n!)^2}|$ as $g(x)$. Checking for absolute convergence I get:
$\lim_{n \to \infty} \left|\dfrac{f_{n+1}(x)}{f_n(x)}\right| = \lim_{n \to \infty}\left|\dfrac{u^{2n+1}}{2^{n+1}\,[(n+1)!]^2} \; \dfrac{2^n\,(n!)^2}{u^n}\right| \, = \, \lim_{n \to \infty} \left|\dfrac{u}{2(n+1)^2}\right| = 0 < 1 \quad , \quad \forall u$
As this sum is absolutely convergent for all $u$, does taking this sum as $g(x)$ imply the interchange of integration and summation in the beginning is justified? And do I have show that this sum as the dominating function $g(x)$ is integrable?
Best Answer
Absolute convergence does not help. To apply above version of DCT you have to note that $e^{-u} \sum\limits_{n=0}^{k} \frac {u^{n}} {2^{n}(n!)^{2}} \leq e^{-u} \sum\limits_{n=0}^{k} \frac {u^{n}} {2^{n}n!}\leq e^{-u} e^{u/2}=e^{-u/2}$ and you can take $g(u)=e^{-u/2}$ in that theorem.