Find the primitive function of $\sum_{n=1}^{\infty}\frac{(-1)^n(2n+2)}{n!}(x-1)^{2n+1}$

Question

Find the primitive function of $\sum_{n=1}^{\infty}\frac{(-1)^n(2n+2)}{n!}(x-1)^{2n+1}$

My attempt:

In order to integrate, I'm trying to find the radius of convergence:

Let $t=(x-1) \Rightarrow \sum_{n=1}^{\infty}\frac{(-1)^n(2n+2)}{n!}t^{2n+1}$

$$ a_n = \begin{cases} \frac{(-1)^k(2k+2)}{k!}, & \text{$n = 2k+1$} \\ 0, & \text{else} \end{cases}$$

$$ \sqrt[n]{|a_n|} = \begin{cases} (\frac{2k+2}{k!})^{\frac{1}{2k+1}}, & \text{$n = 2k+1$} \\ 0, & \text{else} \end{cases}$$

Hence, one of the partial limits is zero and the other:

$$\lim_{k \to \infty}{(\frac{2k+2}{k!})^{\frac{1}{2k+1}}}=0^0=??$$

Answer 1

Note that$$\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}(2n+4)}{(n+1)!}(x-1)^{2n+3}}{\frac{(-1)^n(2n+2)}{n!}(x-1)^{2n+1}}\right|=\lim_{n\to\infty}\frac{2n+4}{2n+2}\times\frac1{n+1}|x-1|^2=0.$$Therefore, the series converges everywhere. A primitive will be$$\sum_{n=1}^\infty\frac{(-1)^n}{n!}(x-1)^{2n+2}.$$