Question

Find the sum of the convergent series. sigma_n = 1^infinity 12/n(n + 2)

Answer 1

Looks like the series is

$\displaystyle\sum_{n=1}^\infty \frac{12}{n(n+2)}$

Split up the summand into partial fractions:

$\dfrac{12}{n(n+2)}=\dfrac6n-\dfrac6{n+2}$

The series has kth partial sum

$S_k=\displaystyle\sum_{n=1}^k\frac{12}{n(n+2)}$

$S_k=\left(6-2\right)+\left(3-\dfrac32\right)+\left(2-\dfrac65\right)+\left(\dfrac32-1\right)+\cdots+\left(\dfrac6{k-2}-\dfrac6k\right)+\left(\left(\dfrac6{k-1}-\dfrac6{k+1}\right)+\left(\dfrac6k-\dfrac6{k+2}\right)$

Several intermediate terms cancel to leave us with

$S_k=6+3-\dfrac6{k+1}-\dfrac6{k+2}$

As $k\to\infty$, the last two terms converge to 0, so that

$\displaystyle\lim_{k\to\infty}S_k=\sum_{n=1}^\infty\frac{12}{n(n+2)}=\boxed{9}$