TITLE: Evaluating the sum : $\;\frac{1}{3}+\frac{1}{4}.\frac{1}{2!}+\frac{1}{5}.\frac{1}{3!}+\ldots$
QUESTION [6 upvotes]: How to evaluate this sum?
$$\frac{1}{3}+\frac{1}{4}.\frac{1}{2!}+\frac{1}{5}.\frac{1}{3!}+\ldots$$
Please give some technique. Binomial not working.

REPLY [17 votes]: Note
\begin{align*}\sum_{n=1}^{\infty}\dfrac{1}{(n+2)n!}&=\sum_{n=1}^{\infty}\dfrac{n+1}{(n+2)!}=\sum_{n=1}^{\infty}\dfrac{(n+2)-1}{(n+2)!}=\sum_{n=1}^{\infty}\left(\dfrac{1}{(n+1)!}-\dfrac{1}{(n+2)!}\right)\\
&=\dfrac{1}{2}-\lim_{n\to\infty}\dfrac{1}{(n+2)!}\\
&=\dfrac{1}{2}
\end{align*}

REPLY [13 votes]: $$\sum_{k=1}^{+\infty}\frac{1}{(k+2)k!} = \int_{0}^{1}\sum_{k=1}^{+\infty}\frac{x^{k+1}}{k!}\,dx=\int_{0}^{1}x\left(e^{x}-1\right)\,dx=\color{red}{\frac{1}{2}}.$$