Closed-form summation of $\sum_{i=1}^n i\frac{x^i}{i!}$

Question

Is there any way to find the closed-form of this finite summation, knowing that x<1? It is part of a larger equation that I am trying to solve/simplify, which has proven to use a lot of theory that I am not yet familiar with. $$\sum_{i=1}^n i\frac{x^i}{i!}$$

If anyone could help it would be highly appreciated.

Answer 1

HINT

$$i \frac{x^i}{i!} = x \frac{d}{dx} \frac{x^{i}}{i!}$$

Can you go on from here? The series now becomes straightforward!

Answer 2

The exponential sum function is defined as: $$e_n(x)=\sum_{k=1}^n \frac{x^k}{k!}$$ Using it's derivative: $$e'_n(x)=\sum_{k=1}^n k\frac{x^{k-1}}{k!}= \sum_{k=1}^n \frac{x^{k-1}}{(k-1)!} = 1 + e_n(x)$$ We have that your series is simply $xe'_n(x)$, or $x+ xe_n(x)$. Unfortunately, the "nicest" representation I could find involves the upper incomplete gamma function: $$e_n(x) = \frac{e^x\Gamma(n+1,x)}{\Gamma(n+1)}$$ Which leads to your expression being:

$$x + \frac{xe^x\Gamma(n+1,x)}{\Gamma(n+1)}$$