Taylor series and calculating a sum with its help

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.

• Well, It seems easier to separate the sum into the sum of $\frac n{2^nn!}$ and $\frac 2{2^nn!}$. – lulu Jun 21 at 13:51
• That's the Taylor series of $f(x-1)$ isn't it? Note that $f(1)=e^{-2}$, so the constant term is incorrect. – saulspatz Jun 21 at 13:58
• How should I fix it? – MathIsTheWayOfLife Jun 21 at 14:06
• Well, you have to actually calculate the Taylor series. I don't know what it is off the top of my head. – saulspatz Jun 21 at 14:34

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 $$x=1$$.