How do you show that $\sum_{n=0}^\infty\frac1{n!(n+2)}=1$?

I've been trying to understand the result of this sum: $$\sum_{n=0}^\infty\frac1{n!(n+2)}=\frac12+\frac13+\frac18+\frac1{30}+\frac1{144}+\dots=1$$ Could you show me how to obtain 1 as result?

• It's a telescoping series. Try computing and simplifying the partial sums into a fraction; you might be able to spot the pattern. – Theo Bendit Jul 17 '18 at 16:41

2 Answers

HINT:

Note that we can write

$$\frac{1}{n!(n+2)}=\frac{(n+2)-1}{(n+2)!}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}$$

Hint Let $f(x)=\sum_{n=0}^\infty \frac{x^n}{n!}=\exp(x)$. Note that $$\sum_{n=0}^\infty\frac{1}{n!(n+2)}=\sum_{n=0}^\infty\frac{1}{n!}\int_{0}^1x^{n+1}\, dx=\int_{0}^1\sum _{n=0}^\infty\frac{x^{n+1}}{n!}\, dx=\int_{0}^1x\exp(x)\, dx.$$ Assumes knowledge of calculus.

• Can you explain how you obtained the integral form? – Szeto Jul 17 '18 at 16:53
• @Szeto notice that $\int_0^1 x^{n+1} = \frac{1}{n+2} x^{n+2} \rvert_0^1 = \frac{1}{n+2}$ – WorldSEnder Jul 17 '18 at 20:23