# Thread: Infinite series of e^x?

1. ## Infinite series of e^x?

Hi,

So, I'm trying to find the sum of:

$\sum_{n=0}^{\infty}\frac{x^n}{(n+1)!}$

I know that $\sum_{n=0}^{\infty}\frac{x^n}{n!} = e^x....$

But How can I use this to find my sum?

Thanks

2. Right, compare the two series...

$\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 + \frac{1}{5!}x^5 + \ldots$

$\sum_{n=0}^{\infty} \frac{x^n}{(n+1)!} = 1 + \frac{1}{2!}x + \frac{1}{3!}x^2 + \frac{1}{4!}x^3 + \frac{1}{5!}x^4 + \ldots$

Compare coefficients and you can see if you divide the first series by x you get the second with the exception of an extra $\frac{1}{x}$ term.

Hence just subtract that and you get...

$\sum_{n=0}^{\infty} \frac{x^n}{(n+1)!} = \frac{e^x - 1}{x}$

3. Originally Posted by Deadstar
Posting solution hold on...

Right, compare the two series...

$\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 + \frac{1}{5!}x^5 + \ldots$

$\sum_{n=0}^{\infty} \frac{x^n}{(n+1)!} = 1 + \frac{1}{2!}x + \frac{1}{3!}x^2 + \frac{1}{4!}x^3 + \frac{1}{5!}x^4 + \ldots$

Compare coefficients and you can see if you divide the first series by x you get the second with the exception of an extra $\frac{1}{x}$ term.

Hence just subtract that and you get...

$\sum_{n=0}^{\infty} \frac{x^n}{(n+1)!} = \frac{e^x - 1}{x}$
Why thank you sir.

4. $\sum\limits_{n\ge 0}{\frac{x^{n}}{(n+1)!}}=\frac{1}{x}\sum\limits_{n \ge 1}{\frac{x^{n}}{n!}}=\frac{1}{x}\left( e^{x}-1 \right).$