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Compute the following sum for any x?

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

I am having trouble to compute that sum. It looks like geometric series but I don't know where to start. Can everyone help me with some hints? Thank you.

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Hints:

$$e^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots$$

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Hint

Consider $$S=\sum_{n=0}^\infty {(x-1)^n\over (n+2)!}=\frac{1}{(x-1)^2}\sum_{n=0}^\infty {(x-1)^{n+2}\over (n+2)!}$$ Adding and substracting the first terms $$(x-1)^2S=\sum_{k=0}^\infty {(x-1)^{k}\over k!}-\cdots$$

I am sure that you can take from here.

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$$\sum_{n=0}^\infty {(x-1)^n\over (n+2)!}={1 \over (x-1)^2} \sum_{n=0}^\infty {(x-1)^{n+2} \over (n+2)!}$$We know the expansion of $e^x$ which is similar to the above expression. Hence $$S={e^{x-1} \over(x-1)^2}$$

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