Convergence of $\sum_{n=1}^\infty\frac{n}{(n+1)!}$ Can someone give an explanation using the definition of convergence in partial sum to show how the above infinite sum converges to 1?
Thanks
 A: This is what's known as a telescoping series. Notice how
$$\frac{n}{(n+1)!} = \frac{(n+1)-1}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}$$
Therefore, since all terms except the first and last cancel out,
$$\sum_{n=1}^k\frac{n}{(n+1)!} = $$$$\left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+...+\left(\frac{1}{(k-1)!}-\frac{1}{k!}\right) + \left(\frac{1}{k!}-\frac{1}{(k+1)!}\right) = $$
$$\frac{1}{1!} - \frac{1}{(k+1)!} = 1 - \frac{1}{(k+1)!}$$
Taking the limit as $k$ goes to $\infty$, we see that the sum converges to $1$. 
A: The answer by florence is nice and probably the best approach. Here's another, just for fun. We have
$$e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$$
This is a power series which is absolutely convergent everywhere, so we're free to manipulate at will.
First subtract $1$ from both sides:
$$e^x - 1 = \sum_{n=1}^{\infty}\frac{x^n}{n!}$$
Divide both sides by $x$ to obtain
$$\frac{e^x - 1}{x} = \sum_{n=1}^{\infty}\frac{x^{n-1}}{n!} = \sum_{n=0}^{\infty}\frac{x^n}{(n+1)!}$$
Differentiate both sides to obtain
$$\frac{xe^x - (e^x - 1)}{x^2} = \sum_{n=1}^{\infty}n\frac{x^{n-1}}{(n+1)!}$$
(We can include or exclude the $n=0$ term as it is zero.)
Finally, evaluate at $x=1$:
$$1 = \sum_{n=1}^{\infty}\frac{n}{(n+1)!}$$
