How do I evaluate this:$\sum_{n=1}^{\infty}\frac{1}{n²}(e^x −1 −\frac{x}{1!} −\frac{x²}{2!}−\cdots\frac{x^n}{n!})$?

Question

How do i evaluate this sum :$$\sum_{n=1}^{\infty}\frac{1}{n²}(e^x −1 −\frac{x}{1!} −\frac{x²}{2!}−\cdots\frac{x^n}{n!})$$

Note: I 'd surprised if it is convergent

Thank you for any help.

Answer 1

The expression in parentheses converges to $0$ because $e^x=\sum\frac{x^n}{n!}$. Together with the fact that $\sum\frac1{n^2}$ converges absolutely, we conclude that your series converges as well.

Answer 2

Fix $x \in \mathbb{R}$. Using the LaGrangian error term, $$e^x = \sum_{k=0}^n \frac{x^k}{k!} + \frac{e^{c_n}\;x^{n+1}}{(n+1)!},$$ for some $c_n \in (0,x)$ or $c_n \in (x,0)$. Noting that $e^c$ is bounded in the interval, the convergence is the same as the convergence of the series \begin{align*} \sum_{n=1}^{\infty}\frac{x^{n+1}}{n^2(n+1)!}, \end{align*} which converges for all $x \in \mathbb{R}$ by the ratio test.