Question

How do you test the series `Sigma 1/((n+1)(n+2))` from n is `[0,oo)` for convergence?

Answer 1

It converges

`sum_(n=0)^oo1/((n+1)(n+2))=1`

Explanation:

`1/((n+1)(n+2))=1/(n+1)-1/(n+2)`

`sum_(n=0)^m1/((n+1)(n+2))=sum_(n=0)^m1/(n+1)-sum_(n=0)^m1/(n+2)`

`=sum_(n=0)^m1/(n+1)-sum_(n=1)^(m+1)1/(n+1)=1-1/(m+2)`

then `lim_(m->oo)sum_(n=0)^m1/((n+1)(n+2))=1-lim_(m->oo)1/(m+2)=1`

Answer 2

`sum_(n=0)^oo 1/((n+1)(n+2))`

is convergent, based on the direct comparison test.

Explanation:

The series:

`sum_(n=0)^oo 1/((n+1)(n+2))`

has positive terms, so we can use the direct comparison test identifying another convergent series:

`sum_(n=0)^oo b_n`

such that:

`1/((n+1)(n+2)) < b_n` for `n > N`.

Now note that clearly, for `n > 1`:

`(n+1) > n => 1/(n+1) < 1/n`

and also:

`(n+2) > n => 1/(n+2) < 1/n`

and multiplying the two inequalities:

`1/((n+1)(n+2)) < 1/n^2` for `n > 1`

Now,

`sum_(n=1)^oo1/n^2 = pi^2/6`

is convergent, so also our series is convergent.