Finding the sum $\sum_{n=0}^\infty\frac{(x+1)^{n+2}}{(n+2)!}$. I cannot use my usual methods that I am use to.

Question

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?

Answer 1

$$\sum_{n=0}^\infty\frac{(x+1)^{n+2}}{(n+2)!}= \sum_{n=2}^\infty\frac{(x+1)^{n}}{(n)!} $$

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

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

Answer 2

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 $D=-2$.