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What is the function representation of this power series?

[Summation from n=0 to infinity of ($x^n)(n+1)!/n!$

The solution is $\frac{1}{(1-x)^-2}$ but how???

I know that $\sum_{n=0}^{\infty}(x^n)/n! = e^x$, but I don't know how to get to the solution from there.

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Notice that $$\frac{(n+1)!}{n!}x^n=\frac{(n+1)n!}{n!}x^n=(n+1)x^n$$ So, $$\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty (n+1)x^n=\sum_{n=0}^\infty nx^n +\sum_{n=0}^\infty x^n=x\sum_{n=0}^\infty nx^{n-1} +\sum_{n=0}^\infty x^n=$$ $$x \frac d {dx}\Big( \sum_{n=0}^\infty x^n\Big)+\sum_{n=0}^\infty x^n$$

I am sure that you cn take it from here.

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  • $\begingroup$ how did you get an extra n! to cancel out the original n! in the denominator? $\endgroup$
    – JavaBeginner
    Apr 6, 2016 at 6:12
  • $\begingroup$ Remember that $k!=k\times (k-1)!=k\times (k-1)\times (k-2)!$ $\endgroup$
    – Claude Leibovici
    Apr 6, 2016 at 6:13
  • $\begingroup$ can you please elaborate a bit more, i'm not so sure what to do with the differentiate summation $\endgroup$
    – JavaBeginner
    Apr 6, 2016 at 6:21
  • $\begingroup$ What is $\sum_{n=0}^\infty x^n$ ? $\endgroup$
    – Claude Leibovici
    Apr 6, 2016 at 6:23
  • $\begingroup$ lol i got it , thanks $\endgroup$
    – JavaBeginner
    Apr 6, 2016 at 6:28

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