Prove $\sum_{i=0}^\infty \frac{1}{(i+2)i!}=1$

Question

How can I prove that $$\sum_{i=0}^\infty \frac{1}{(i+2)i!}=1$$? Does it relate to the exponential function?

Answer 1

You can rewrite the problem as follows: $$\sum_{i=0}^{\infty}\frac{1}{(i+2)\cdot i!}=\sum_{i=0}^{\infty}\frac{i+1}{(i+2)!}=\sum_{i=0}^{\infty}\left (\frac{1}{(i+1)!}-\frac{1}{(i+2)!}\right ),$$

which is an obvious example of a telescoping series. Therefore, this yields the expected sum $1$.

Answer 2

Hint

$$xe^x=\sum_{i=0}^{\infty}\frac{x^{i+1}}{i!}$$ $$\int_{0}^{1}xe^xdx=\sum_{i=0}^{\infty}\frac{1}{(i+2)i!}$$

Answer 3

You can relate the sum back to the Taylor series for the exponential function centered at $x=0$. That is, $$\frac{1}{(i+2)i!} = \frac{i+1}{(i+2)!} = \left(\frac{d}{dx}\frac{x^{i+1}}{(i+2)!}\right) \mbox{ evaluated at } x=1.$$ Then you can relate $$\sum_{i=0}^\infty \frac{x^{i+1}}{(i+2)!}$$ back to the exponential function’s Taylor expansion and put all the pieces together.