Evaluating the sum : $\;\frac{1}{3}+\frac{1}{4}.\frac{1}{2!}+\frac{1}{5}.\frac{1}{3!}+\ldots$

Question

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.

Answer 1

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*}

Answer 2

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

Answer 3

Some details regarding convergence aside, you have $$e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$$ so $$x e^x = \sum_{k=0}^\infty \frac{x^{k+1}}{k!}$$ and $$ \int_0^t x e^x \, dx = \sum_{k=0}^\infty \frac{1}{k+2} \frac{t^{k+2}}{k!}.$$ Thus $$\sum_{k=1}^\infty \frac{1}{k+2} \frac{1}{k!} = \int_0^1 x e^x \, dx - 1.$$ The last integral is easy to evaluate.