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 (12)/(n(n+2))`

Split up the summand into partial fractions:

`(12)/(n(n+2))=\frac6n-\frac6{n+2}`

The series has kth partial sum

`S_k=\displaystyle\sum_(n=1)^k(12)/(n(n+2))`

Several intermediate terms cancel to leave us with

`S_k=6+3-\frac6{k+1}-\frac6{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(12)/(n(n+2))=\boxed{9}`