So this function is given:
$$f(x)=xe^{-2x}$$
I have to calculate its Taylor series around $x=1$ and use it in some way to calculate the following sum:
$$\sum_{n=1}^{\infty} \frac{n+2}{(2n)!!}$$
I do know the fact: $$(2n)!!=2^nn!$$ $$\sum_{n=1}^{\infty} \frac{n+2}{(2n)!!}=\sum_{n=1}^{\infty} \frac{n+2}{2^nn!}$$
And after 1st step of taylor series i get this:
$$f(x)=(x-1)\sum_{n=0}^{\infty} \frac{(-2(x-1))^n}{n!}= \sum_{n=0}^{\infty} \frac{(-1)^n2^n(x-1)^{n+1}}{n!} = -1 + \sum_{n=1}^{\infty} \frac{(-1)^{n}2^n(x-1)^{n+1}}{n!} $$
But I have no idea what to do next. Any help would be appreciated.
HINT
To calculate the Taylor series of $f$, start by calculating the Taylor series of $g(x)=e^{-2x}$. Obviously, $g^{(n)}(x)=(-2)^ne^{-2x}$ so $g^{(n)}(1)=(-2)^ne^{-2}$ and $$g(x)=\sum_{n=0}^\infty(-2)^ne^{-2}(x-1)^n$$ is the Taylor series of $g$ about $x=1$.
Now $f(x)=xg(x)=(1+(x-1))g(x)$, so you can easily get the Taylor series for $f$ about $x=1$.