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

Feb 10, 2017

It converges

${\sum}_{n = 0}^{\infty} \frac{1}{\left(n + 1\right) \left(n + 2\right)} = 1$

#### Explanation:

$\frac{1}{\left(n + 1\right) \left(n + 2\right)} = \frac{1}{n + 1} - \frac{1}{n + 2}$

${\sum}_{n = 0}^{m} \frac{1}{\left(n + 1\right) \left(n + 2\right)} = {\sum}_{n = 0}^{m} \frac{1}{n + 1} - {\sum}_{n = 0}^{m} \frac{1}{n + 2}$

$= {\sum}_{n = 0}^{m} \frac{1}{n + 1} - {\sum}_{n = 1}^{m + 1} \frac{1}{n + 1} = 1 - \frac{1}{m + 2}$

then ${\lim}_{m \to \infty} {\sum}_{n = 0}^{m} \frac{1}{\left(n + 1\right) \left(n + 2\right)} = 1 - {\lim}_{m \to \infty} \frac{1}{m + 2} = 1$

Feb 10, 2017

${\sum}_{n = 0}^{\infty} \frac{1}{\left(n + 1\right) \left(n + 2\right)}$

is convergent, based on the direct comparison test.

#### Explanation:

The series:

${\sum}_{n = 0}^{\infty} \frac{1}{\left(n + 1\right) \left(n + 2\right)}$

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

${\sum}_{n = 0}^{\infty} {b}_{n}$

such that:

$\frac{1}{\left(n + 1\right) \left(n + 2\right)} < {b}_{n}$ for $n > N$.

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

$\left(n + 1\right) > n \implies \frac{1}{n + 1} < \frac{1}{n}$

and also:

$\left(n + 2\right) > n \implies \frac{1}{n + 2} < \frac{1}{n}$

and multiplying the two inequalities:

$\frac{1}{\left(n + 1\right) \left(n + 2\right)} < \frac{1}{n} ^ 2$ for $n > 1$

Now,

${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 2 = {\pi}^{2} / 6$

is convergent, so also our series is convergent.