Find the sum of the series n/(n+1)! for n from 1 to infinity.

Step-by-step solution

1. 1

   Split the numerator so it matches the factorial below it. The obstacle is that $n$ and $(n+1)!$ are out of step. Writing $n = (n+1) - 1$ fixes that:

   $\frac{n}{(n+1)!} = \frac{(n+1) - 1}{(n+1)!} = \frac{n+1}{(n+1)!} - \frac{1}{(n+1)!}$

2. 2

   Simplify the first piece. Because $(n+1)! = (n+1)\cdot n!$, the factor $(n+1)$ cancels:

   $\frac{n+1}{(n+1)!} = \frac{1}{n!}$

   so the general term becomes a difference of consecutive reciprocal factorials:

   $\frac{n}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}$

3. 3

   Write the partial sum and watch the middle cancel. With $a_n = \tfrac{1}{n!}$ the term is $a_n - a_{n+1}$, so

   $S_N = \sum_{n=1}^{N}\left(\frac{1}{n!} - \frac{1}{(n+1)!}\right) = \frac{1}{1!} - \frac{1}{(N+1)!}$

   Every interior term appears once with each sign. This is what makes the series telescoping rather than merely convergent.

4. 4

   Take the limit. Factorials grow faster than any power, so $\frac{1}{(N+1)!} \to 0$ and

   $\sum_{n=1}^{\infty} \frac{n}{(n+1)!} = \lim_{N\to\infty} S_N = 1$

5. 5

   Sanity-check numerically. The first five terms are $\tfrac12 + \tfrac{2}{6} + \tfrac{3}{24} + \tfrac{4}{120} + \tfrac{5}{720} = 0.5 + 0.333333 + 0.125 + 0.033333 + 0.006944 = 0.998611$, and $1 - \tfrac{1}{6!} = 1 - 0.001389 = 0.998611$ exactly, matching the closed form for $S_5$.