Hello i am trying this exercise:
Use the MacLaurin series of the function $f(x)=xe^{-x^2}$ to prove that
$$\sum_{n=1}^\infty (-1)^{n+1}\frac{2n+1}{2^{n}n!} =1$$
To prove this i differentiated the series
$$f(x)=x\sum_{n=0}^\infty (-1)^{n}\frac{x^{2n}}{n!}=\sum_{n=0}^\infty (-1)^{n}\frac{x^{2n+1}}{n!}$$
$$f'(x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}(2n-1)}{(n-1)!}x^{2n-2}$$
but i don't see how to use that $f'(0)=1$ to prove that. I rewrote the expression as
$$1=\sum_{n=1}^\infty (-1)^{n+1}\frac{2n+1}{2^{n}n!}=\sum_{n=1}^\infty (-1)^{n+1}\frac{2n+1}{n!} (\frac{1}{2})^{n} =1$$
but that is as far i go, can anyone help me to see what i am doing wrong?
The term-wise derivative of $$ f(x)=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{n!}$$ is $$ f'(x)=\sum_{n=0}^\infty (-1)^n\frac{(2n+1)x^{2n}}{n!}.$$ I think you want to have a look at $f'(1/\sqrt 2)$.