# Find the sum of the following infinite series $e^{-x}\sum_{i=0}^{\infty}\frac{i.x^i}{i!}$

Find the sum of the following infinite series $$e^{-x}\sum_{i=0}^{\infty}\dfrac{i.x^i}{i!}$$

The summation looks like an exponential series but how to tackle that?$$0+\frac{x}{1!}+\frac{2x^2}{2!}+...$$

$$\sum_{n=0}^{\infty}\dfrac{nx^n}{n!} = x\cdot\sum_{n=1}^{\infty}\dfrac{x^{n-1}}{(n-1)!} = xe^x.$$
So $e^{-x}$ times that is just $x$. Is this what you're asking?
Note that $$\sum_{i=0}^\infty \frac{ix^i}{i!} =xD(e^x)=xe^x.$$ So $$e^{-x}\left(\sum_{i=0}^\infty \frac{ix^i}{i!}\right) =e^{-x}xe^x =x.$$