Determining sum of series

Question

I am supposed to solve the following problem:

Find sum of the series and range of convergence of: $\sum_{n=0}^{\infty}\frac{2n+1}{n!}x^{2n}$

I know that I have to use Taylor polynomial for $e^x$ because of factorial. But I do not have idea, how to do it. Can anyone help?

Answer 1

Hint : \begin{eqnarray*} \sum_{n=0}^{\infty}\frac{2n+1}{n!}x^{2n} &=& 2\sum_{\color{red}{n=1}}^{\infty}\frac{x^{2n}}{(n-1)!} + \sum_{n=0}^{\infty}\frac{x^{2n}}{n!}. \\ \end{eqnarray*} Now use \begin{eqnarray*} e^{x^2}= \sum_{n=0}^{\infty}\frac{x^{2n}}{n!}. \end{eqnarray*}

$e^{x^2}+2 x^2 (e^{x^2}-1)$.

Answer 2

Integrating term by term (radius of convergence is infinity so this is fine) we get $$ \sum_{n=0}^{\infty} \frac{1}{n!} x^{2n+1} = x \, \sum_{n=0}^{\infty} \frac{1}{n!} (x^2)^n = xe^{x^2}\,. $$ Now we differentiate to work out the original function, giving $$ (1+ 2x^2)e^{x^2}\,. $$

Answer 3

Hint: split it in two: \begin{align} \sum_{n=0}^{\infty}\frac{2n+1}{n!}x^{2n}&=\sum_{n=0}^{\infty}2n\,\frac{ x^{2n}}{n!}+\sum_{n=0}^{\infty}\frac{\bigl(x^2\bigr)^n}{n!}=x\sum_{n=0}^{\infty} \frac{2n\,x^{2 n-1}}{n!}+\sum_{n=0}^{\infty}\frac{\bigl(x^2\bigr)^n}{n!} \\ &=x\biggl(\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}\biggr)^{\!\prime}+\sum_{n=0}^{\infty}\frac{\bigl(x^2\bigr)^n}{n!}. \end{align}