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

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

any help appreciated, Thanks.

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

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

any help appreciated, Thanks. - Nov 23rd 2009, 11:11 PMProve It
Are you trying to prove it converges absolutely?

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

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

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

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

What happens as $\displaystyle n \to \infty$? - Nov 23rd 2009, 11:20 PMItsygo
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 AMHallsofIvy
- Nov 24th 2009, 07:57 AMDrexel28
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 $\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}$ 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 $\displaystyle f'(x)=\sum_{n=1}^{\infty}\frac{2nx^{2n-1}}{n!}=\sum_{n=0}^{\infty}\frac{2(n+1)x^{2(n+1)-1}}{(n+1)!}=\sum_{n=0}^{\infty}\frac{2x^{2n+1}}{n! }=2xf(x)$. Therefore, $\displaystyle f'(x)=2xf(x)\implies \frac{f'(x)}{f(x)}=2x\implies \ln\left(f(x)\right)=x^2+C\implies f(x)=C_1e^{x^2}$. And seeing that $\displaystyle f(0)=1$ we can conclude that $\displaystyle C_1=1\implies f(x)=e^{x^2}$ therefore $\displaystyle \sum_{n=0}^{\infty}\frac{x^{2n+1}}{n!}=xf(x)=xe^{x ^2}$.