Taylor Series Expansion to Find Value of Series

Question

How to use Taylor series of $xe^x$ to prove that $\sum_{n=0}^\infty\frac1{(n+2)n!}=1$?

Answer 1

The Taylor series expansion of $xe^x$ is $$xe^x=\sum_{k=0}^\infty\frac{x^{k+1}}{k!}$$ Integrating both sides with respect to $x$ gives $$\sum_{k=0}^\infty\frac{x^{k+2}}{k!(k+2)}=xe^x-e^x+C$$ When $x=0$ the LHS sum is clearly $0$ hence we can find $C$ $$x=0\implies0=0-1+C\implies C=1$$ $$\therefore\sum_{k=0}^\infty\frac{x^{k+2}}{k!(k+2)}=xe^x-e^x+1$$ Then plugging $x=1$ into both sides of the equation gives $$\sum_{k=0}^\infty\frac{1}{k!(k+2)}=1\cdot e^1-e^1+1=1$$