Find a function that is equivalent to the series $\sum_{n=0}^\infty\frac{(x/2)^n}{(n+1)!}$

Question

I need to find a function that is equivalent to $\sum_{n=0}^\infty\frac{(x/2)^n}{(n+1)!}$ I integrated the function and then compared the integral to the known power series $e^x$ by replacing the $x$ and then derived that but I got a very long function in the end that definitely does not look right. Here is a picture of all my work: (The final iteration of the equation is hypothetically equivalent to the given series I just forgot to write that on the paper)

So I need to know:

1) Is my work correct

2) Is there a better way to turn this series into a function

3) If there is a better way, what is the better way

4) What is a function that is equivalent to the given series

Answer 1

You can't get an $n$-dependent result because $n$ is a dummy variable. You want $$\sum_{n\ge0}\frac{(x/2)^n}{(n+1)!}=\frac{2}{x}\sum_{n\ge0}\frac{(x/2)^{n+1}}{(n+1)!}=\frac{2}{x}\sum_{m\ge1}\frac{(x/2)^m}{m!}=\frac{2}{x}\left(e^{x/2}-1\right).$$