Summation Value of Combination Function

Question

$$\sum_{n=1}^{\infty}\frac{5^n}{n!2^n}$$

I know how to prove that the above series is convergent but I'm not sure how to get the value that it converges to. If it were just the geometric series, I could get an end value but the factorial in the bottom complicates matters. Thank you very much.

This is NOT a homework sum. I genuinely do not understand how to solve it.

Answer 1

Hint:

Note that the sum is equivalent to

$$\sum_{n = 1}^{\infty} \frac{(5/2)^n}{n!}$$

Recall the Taylor series for the exponential function:

$$e^x = \sum_{n = 0}^{\infty} \frac{x^n}{n!}$$

The only difference is that the second sum starts at $1$, rather than $0$. Do you see how to find the value now?

Answer 2

$\displaystyle \sum_{n=1}^\infty \dfrac{5^n}{n!\cdot 2^n}$ = $\displaystyle \sum_{n=1}^\infty \dfrac{(\frac{5}{2})^n}{n!}$ = $e^{\frac{5}{2}} - 1$