Strategy for Series — Question 6

*Question 6 Analyze $\displaystyle\sum_{n=1}^{\infty}\frac{n!}{(n+2)!}$.

1. Simplify the factorial quotient.

2. Decide whether telescoping or comparison is more informative.

3. Determine convergence and, if possible, the exact sum.

*Question 6 – Solution

$$\frac{n!}{(n+2)!}=\frac1{(n+1)(n+2)}.$$ This already resembles a telescoping difference.

$$\frac1{(n+1)(n+2)}=\frac1{n+1}-\frac1{n+2}.$$ Thus $$\begin{align*} S_N&=\sum_{n=1}^N\left(\frac1{n+1}-\frac1{n+2}\right)\\&=\frac12-\frac1{N+2}. \end{align*}$$

$$\sum_{n=1}^{\infty}\frac{n!}{(n+2)!}=\lim_{N\to\infty}S_N=\boxed{\frac12}.$$

Comparison with $1/n^2$ proves convergence, but telescoping is better because it also gives the exact sum.