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I am asked to find the sum of the series $$\sum_{n=0}^\infty\frac{(x+1)^{n+2}}{(n+2)!}$$

For some reason (that I don't understand) I can't apply the techniques for finding the sum of the series that I usually would to this one? I think the others that I have done have been geometric series so I could just find the first few terms etc...

This one is really difficult for me to figure out.

As far as my research into this has taken me, it has something to do with writing the series out as a power series about x=a and somehow using this to find a? My teacher told me to take a few derivatives of this? Does she mean to take the derivatives of taylor series or something like that?

I am pretty confused as to the method I need to complete this question and online sources and class lecture information has taken me nowhere but back to the stuff I understand that applies to geometric series. What kind of series is this? How do I find its sum?

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    $\begingroup$ Hint: Consider the Taylor series for $e^x$. $\endgroup$
    – John Douma
    Oct 29 '18 at 3:37
  • $\begingroup$ Another hint:$\sum_{n=0}^\infty a_{n+1}=\sum_{n=0}^\infty a_n-a_0$ $\endgroup$ Oct 29 '18 at 3:44
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$$\sum_{n=0}^\infty\frac{(x+1)^{n+2}}{(n+2)!}= \sum_{n=2}^\infty\frac{(x+1)^{n}}{(n)!} $$

$$=\sum_{n=0}^\infty\frac{(x+1)^{n}}{(n)!} -1-(x+1) $$

$$=e^{x+1}-x-2$$

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  • $\begingroup$ Thanks for replying! I get what you have done other than why after the general taylor representation there is $$-1-(x+1)$$ ? $\endgroup$
    – User1997
    Oct 29 '18 at 4:34
  • $\begingroup$ @KaylaMartin These are the first two terms of the series for $e^{x+1}$ . Since the series was from $n=2$ instead of $n=0$ we need to subtract the first two terms. $\endgroup$ Oct 29 '18 at 6:51
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What your teacher was meaning is that $$\frac d{dx} \sum_{n=0}^\infty\frac{(x+1)^{n+2}}{(n+2)!}= \sum_{n=0}^\infty\frac{(x+1)^{n+1}}{(n+1)!}\tag 1$$ $$\frac {d^2}{dx^2} \sum_{n=0}^\infty\frac{(x+1)^{n+2}}{(n+2)!}= \sum_{n=0}^\infty\frac{(x+1)^{n}}{(n)!}=e^{x+1}\tag 2$$ Integrate both sides of $(2)$ to get $$\int\frac {d^2}{dx^2} \sum_{n=0}^\infty\frac{(x+1)^{n+2}}{(n+2)!}=e^{x+1}+C$$ Make $x=-1$ on both sides to get $C=-1$. So $$\frac d{dx} \sum_{n=0}^\infty\frac{(x+1)^{n+2}}{(n+2)!}= \sum_{n=0}^\infty\frac{(x+1)^{n+1}}{(n+1)!}=e^{x+1}-1\tag 3$$ Integrate both sides of $(3)$ to get $$\sum_{n=0}^\infty\frac{(x+1)^{n+2}}{(n+2)!}=e^{x+1}-x+D$$ Make $x=-1$ on both sides to get $D=-2$.

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  • $\begingroup$ Thanks for your reply! Why is it equal to $e^{x+1}$ ? $\endgroup$
    – User1997
    Oct 29 '18 at 4:52
  • $\begingroup$ @KaylaMartin. If you prefer some thing even more obvious, start using $x+1=y$ $\endgroup$ Oct 29 '18 at 4:55

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