# How do you test the series sum_(n=0)^(oo) n/((n+1)(n+2)) for convergence?

Nov 16, 2017

The series diverge

#### Explanation:

Perform the limit comparison test

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

and ${b}_{n} = \frac{1}{n}$, this series diverge

${a}_{n} > 0$ and ${b}_{n} > 0$, $\forall n \in \mathbb{N}$

${\lim}_{n \to \infty} {a}_{n} / {b}_{n} = {\lim}_{n \to \infty} \left(\frac{\frac{n}{\left(n + 1\right) \left(n + 2\right)}}{\frac{1}{n}}\right)$

$= {\lim}_{n \to \infty} \left({n}^{2} / \left(\left(n + 1\right) \left(n + 2\right)\right)\right)$

$= {\lim}_{n \to \infty} \left({n}^{2} / \left(\left({n}^{2} + 3 n + 2\right)\right)\right)$

$= {\lim}_{n \to \infty} \left(\frac{1}{\left(1 + \frac{3}{n} + \frac{2}{n} ^ 2\right)}\right)$

$= 1$

We conclude that, by the limit comparison test that the series ${a}_{n}$ diverge

Nov 16, 2017

The series:

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

is divergent.

#### Explanation:

The series has only positive terms, so we can use the limit comparison test to compare it with the harmonic series:

${\lim}_{n \to \infty} \frac{\frac{n}{\left(n + 1\right) \left(n + 2\right)}}{\frac{1}{n}} = {\lim}_{n \to \infty} {n}^{2} / \left({n}^{2} + 3 n + 2\right) = 1$

As the limit is finite and positive the two series have the same character, and we know the harmonic series to be divergent, thus also the series:

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

is divergent.

Nov 16, 2017

We can use the integral test to find it diverges.

#### Explanation:

Using the integral test, we find:

$\int \frac{x}{\left(x + 1\right) \left(x + 2\right)} \mathrm{dx} = \int \left(\frac{2}{x + 2} - \frac{1}{x + 1}\right) \mathrm{dx}$

$\textcolor{w h i t e}{\int \frac{x}{\left(x + 1\right) \left(x + 2\right)} \mathrm{dx}} = 2 \ln \left\mid x + 2 \right\mid - \ln \left\mid x + 1 \right\mid + C$

$\textcolor{w h i t e}{\int \frac{x}{\left(x + 1\right) \left(x + 2\right)} \mathrm{dx}} = \ln \left({\left\mid x + 2 \right\mid}^{2} / \left\mid x + 1 \right\mid\right) + C$

$\textcolor{w h i t e}{\int \frac{x}{\left(x + 1\right) \left(x + 2\right)} \mathrm{dx}} > \ln \left({\left\mid x + 1 \right\mid}^{2} / \left\mid x + 1 \right\mid\right) + C = \ln \left(\left\mid x + 1 \right\mid\right) + C$

$\textcolor{w h i t e}{\int \frac{x}{\left(x + 1\right) \left(x + 2\right)} \mathrm{dx}} \to \infty \text{ }$ as $x \to \infty$

So:

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

Nov 16, 2017

See below.

#### Explanation:

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

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

Now considering

$H = {\sum}_{n = 1}^{n} \frac{1}{n}$ we have

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

but as we know $H$ is the so called harmonic series

know as divergent.

Resuming

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