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.
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 $\displaystyle n \to \infty$?
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 $\displaystyle \sum_{n=0}^{\infty}\frac{x^{2n+1}}{n!}=xf(x)=xe^{x ^2}$.