How would I go about solving $$\sum_{n=1}^{\infty} \frac{(n-1)x^{n-1}}{(n-1)!}$$ So far I have tried to of course consider the exponential power series, but I seem to get negative factorials.

  • $\begingroup$ You won’t get negative factorials if you remove the first term, though. Note $1-1=0$ at $n=1$. Alternatively you can use the convention $(-n)!=\infty$ so $1/(-n)!=0$ for $n\ge1$ an integer. This is correct in some sense. $\endgroup$
    – FShrike
    Nov 14, 2023 at 13:27
  • 1
    $\begingroup$ This is also $\sum\limits_{n=0}^{\infty} n \frac{x^n}{n!}$ $\endgroup$
    – Henry
    Nov 15, 2023 at 10:25
  • $\begingroup$ It would have been nice if you had included your best attempt. $\endgroup$
    – Teepeemm
    Nov 15, 2023 at 16:54

1 Answer 1


Note that

\begin{align*} \sum_{n=1}^\infty\frac{(n-1)x^{n-1}}{(n-1)!} &=\sum_{n=2}^\infty\frac{x^{n-1}}{(n-2)!} \\ &=x\sum_{n=2}^\infty\frac{x^{n-2}}{(n-2)!} \\ &=x\sum_{k=0}^\infty\frac{x^k}{k!} \\ &=xe^x. \end{align*}


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