Convergence of $\sum_{n=1}^\infty\frac{n}{(n+1)!}$

Solution 1:

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 $1$.

Solution 2:

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