Originally Posted by
Deadstar Posting solution hold on...
Right, compare the two series...
$\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 + \frac{1}{5!}x^5 + \ldots$
$\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{(n+1)!} = 1 + \frac{1}{2!}x + \frac{1}{3!}x^2 + \frac{1}{4!}x^3 + \frac{1}{5!}x^4 + \ldots$
Compare coefficients and you can see if you divide the first series by x you get the second with the exception of an extra $\displaystyle \frac{1}{x}$ term.
Hence just subtract that and you get...
$\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{(n+1)!} = \frac{e^x - 1}{x}$