Find the Maclaurin series of $$f(x)=xe^x$$ Integrate this series term by term in the closed interval $[0,1]$ and demonstrate that: $$\sum^\infty_{2} \frac{1}{(n-2){} !n} = 1$$
I tried it:
Formula for Maclaurin series: $$f(x)=f(0)+xf'(0)+\frac{x^2}{2 !}f''(0)+...+\frac{x^n}{n !}f^{(n)}(0)+...$$ Therefore I have the following data: $$f(x)=xe^x$$ $$f(0)=xe^0=0$$ $$f'(x)=e^x+xe^x\implies f'(0)=1$$ $$f'(x)=e^x+e^x+xe^x=2e^x+xe^x\implies f''(0)=2$$ $$...$$
Hints:
$$xe^x=x\sum_{k=0}^\infty\frac{x^k}{k!}=\sum_{k=0}^\infty\frac{x^{k+1}}{k!}\implies$$
$$\int_0^1xe^xdx=\sum_{k=0}^\infty\frac1{k!}\int_0^1x^{k+1}dx=\sum_{k=0}^\infty\frac1{k!(k+2)}$$
Now do by parts the left integral.