Calculating the sum $\sum_{n=1}^\infty \frac{n}{(n-1)!}x^n$?

Question

So I've got the sum $$\sum_{n=1}^\infty \frac{n}{(n-1)!}x^n$$

To show that it converges for all real numbers, I used the ratio test. And found the convergence radius to be $$R = \frac{1}{L}, \qquad R = \infty$$ The next task is to calculate the sum, and I feel sort of lost.. I think I want the sum too look like a geometric series. Or substitute it with something else.

Answer 1

Recall that $$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}.$$ First way. Note that $$xe^x=x(e^x)'=x\sum_{n=0}^\infty n\frac{x^{n-1}}{n!}=\sum_{n=1}^\infty \frac{x^{n}}{(n-1)!}.$$ Try to differentiate again and compare the result with your series.

Second way. we have that $$\sum_{n=1}^\infty \frac{n}{(n-1)!}x^n=\sum_{m=0}^\infty \frac{m+1}{m!}x^{m+1}=x^2\sum_{m=1}^\infty \frac{m}{m!}x^{m-1}+x\sum_{m=0}^\infty \frac{x^{m}}{m!}.$$ What then?

Answer 2

Hint:

Use $n=(n-1)+1$ and simplify.

Answer 3

$$e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}=\sum _{n=1}^{\infty }{\frac {x^{n-1}}{(n-1)!}}$$ or $$xe^{x}=\sum _{n=1}^{\infty }{\frac {x^{n}}{(n-1)!}}$$