What is the value of $\sum_{n=0}^{\infty}\frac{(n+1)x^n}{n!}$? I need to find the value of the series $\sum_{n=0}^{\infty}\frac{(n+1)x^n}{n!}$.I've computed its radius of convergence which comes out to be zero.
I'm not getting how to make adjustments in the general terms of the series to get the desired result...
 A: Hint
$$\dfrac{(n+1)x^n}{n!}=x\cdot\dfrac{x^{n-1}}{(n-1)!}+\dfrac{x^n}{n!}$$
Now $$e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$$
A: Let $a_n$ the sequence in the series.
Now $a_n=\frac{z^n}{(n-1)!}+\frac{z^n}{n!}$
Now $\sum_{n=0}^{\infty}\frac{nz^n}{n!}=\sum_{n=1}^{\infty}\frac{z^n}{(n-1)!}=z\sum_{k=0}^{\infty}\frac{z^k}{k!}=ze^z$
Also $\sum_{k=0}^{\infty}\frac{z^k}{k!}=e^z$
So the whole sum is $(z+1)e^z$
A: Is it not so that you can split the sum into two and simplify:
$$\sum_{n=0}^\infty \frac{(n+1)x^n}{n!} = \sum_{n=0}^\infty \frac{nx^n}{n!} + \sum_{n=0}^\infty \frac{x^n}{n!} = \sum_{n=1}^\infty \frac{x^{n}}{(n-1)!} + \sum_{n=0}^\infty \frac{x^n}{n!}= \ldots$$
Considering what reindexing the first sum to start at 0 is, and knowing the expansion for $x \mapsto e^x$ will see you well!
A: $$y=\sum_{n=0}^{\infty}\frac{nx^n}{n!}+\sum_{n=0}^{\infty}\frac{x^n}{n!}$$
$$y=e^x+\sum_{n=0}^{\infty}\frac{nx^n}{n!}$$
$$\frac{y}{x}=\frac{e^x}{x}+\sum_{n=0}^{\infty}\frac{nx^{n-1}}{n!}$$
$$\frac{y}{x}=\frac{e^x}{x}+e^x$$
so
$$y=e^x+xe^x$$
