# Power Series

• Nov 23rd 2009, 11:05 PM
Itsygo
Power Series
Im having trouble with this power series, I know it converges absolutely but im having trouble getting there.

$\sum_{n=0}^{\infty}\frac{x^{2n+1}}{n!}$

any help appreciated, Thanks.
• Nov 23rd 2009, 11:11 PM
Prove It
Quote:

Originally Posted by Itsygo
Im having trouble with this power series, I know it converges absolutely but im having trouble getting there.

$\sum_{n=0}^{\infty}\frac{x^{2n+1}}{n!}$

any help appreciated, Thanks.

Are you trying to prove it converges absolutely?

If so, you need to show that $\lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right| < 1$.

So in your case $a_n = \frac{x^{2n + 1}}{n!}$ and $a_{n + 1} = \frac{x^{2n + 3}}{(n + 1)!}$.

Thus $\left|\frac{a_{n + 1}}{a_n}\right| = \left|\frac{x^{2n + 3}}{(n + 1)!}\cdot \frac{n!}{x^{2n + 1}}\right|$

$= \frac{x^2}{n + 1}$.

What happens as $n \to \infty$?
• Nov 23rd 2009, 11:20 PM
Itsygo
awesome thanks for the quick post, ya sorry I needed to prove a.c.

it eventually gets infinity small < 1
• Nov 24th 2009, 03:42 AM
HallsofIvy
Quote:

Originally Posted by Itsygo
Im having trouble with this power series, I know it converges absolutely but im having trouble getting there.

$\sum_{n=0}^{\infty}\frac{x^{2n+1}}{n!}$

any help appreciated, Thanks.

In fact, you can determine what that sum is equal to. It is well known that $e^x= \sum_{n=0}^\infty \frac{x^n}{n!}$. Here you have $\sum_{n=0}^\infty \frac{x^{2n+1}}{n!}= \sum_{n=0}^\infty x\frac{(x^2)^n}{n!}= x e^{x^2}$
• Nov 24th 2009, 07:57 AM
Drexel28
Quote:

Originally Posted by HallsofIvy
In fact, you can determine what that sum is equal to. It is well known that $e^x= \sum_{n=0}^\infty \frac{x^n}{n!}$. Here you have $\sum_{n=0}^\infty \frac{x^{2n+1}}{n!}= \sum_{n=0}^\infty x\frac{(x^2)^n}{n!}= x e^{x^2}$

If this power series isn't known to you (which it really should be) we can find it slighly differently. Suppose we just want to find the explicity function described by $f(x)=\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}$ notice that we may (take for granted here that you can...there is criterion it needs to meet but it does...so) differentate to obtain $f'(x)=\sum_{n=1}^{\infty}\frac{2nx^{2n-1}}{n!}=\sum_{n=0}^{\infty}\frac{2(n+1)x^{2(n+1)-1}}{(n+1)!}=\sum_{n=0}^{\infty}\frac{2x^{2n+1}}{n! }=2xf(x)$. Therefore, $f'(x)=2xf(x)\implies \frac{f'(x)}{f(x)}=2x\implies \ln\left(f(x)\right)=x^2+C\implies f(x)=C_1e^{x^2}$. And seeing that $f(0)=1$ we can conclude that $C_1=1\implies f(x)=e^{x^2}$ therefore $\sum_{n=0}^{\infty}\frac{x^{2n+1}}{n!}=xf(x)=xe^{x ^2}$.