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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.

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

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