# How do you test the series Sigma sqrtn/(3n+2) from n is [0,oo) for convergence?

Mar 1, 2017

See below.

#### Explanation:

We can use the Integral Test.

This means we consider the analogous function $f \left(n\right) = {a}_{n}$, and we test the limit:

${\int}_{1}^{\infty} f \left(x\right) \setminus \mathrm{dx} = {\lim}_{t \to \infty} {\int}_{1}^{t} f \left(x\right) \setminus \mathrm{dx} < \infty$

In this case:

${\lim}_{t \to \infty} {\int}_{1}^{t} \frac{\sqrt{x}}{3 x + 2} \setminus \mathrm{dx} < \infty$

With sub: $z = \sqrt{x} , {z}^{2} = x , 2 z \mathrm{dz} = \mathrm{dx}$, we have:

$= {\lim}_{t \to \infty} {\int}_{1}^{t} \frac{2 {z}^{2}}{3 {z}^{2} + 2} \setminus \mathrm{dz}$

$= {\lim}_{t \to \infty} {\int}_{1}^{t} \frac{\left(\frac{2}{3}\right) \left(3 {z}^{2} + 2 - 2\right)}{3 {z}^{2} + 2} \setminus \mathrm{dz}$

$= {\lim}_{t \to \infty} {\int}_{1}^{t} \frac{2}{3} - \textcolor{red}{\frac{4}{9 {z}^{2} + 6}} \setminus \mathrm{dz} q \quad \triangle$

If we focus on the indefinite integral for the red term, ie:

$= \int \frac{4}{9 {z}^{2} + 6} \setminus \mathrm{dz}$

... we can use the sub: $9 {z}^{2} = 6 {\tan}^{2} \theta , 3 z = \sqrt{6} \tan \theta , 3 \mathrm{dz} = \sqrt{6} {\sec}^{2} \theta d \theta$. This leads to:

$= \int \frac{4}{6 {\tan}^{2} \theta + 6} \setminus \frac{\sqrt{6}}{3} {\sec}^{2} \theta \setminus d \theta$

$= \frac{4}{3 \sqrt{6}} \int \setminus d \theta$

$= \frac{4}{3 \sqrt{6}} \arctan \left(\frac{3 z}{\sqrt{6}}\right) + C$

Putting $\triangle$ all together yields:

${\lim}_{t \to \infty} {\left[\frac{2}{3} \sqrt{x} - \frac{4}{3 \sqrt{6}} \arctan \left(\sqrt{\frac{3 x}{2}}\right)\right]}_{1}^{t}$

Now we know that: ${\lim}_{\phi \to \infty} \arctan \left(\phi\right) = \frac{\pi}{2}$ so the leading $\sqrt{x}$ term dominates the limit, and means that this integral does not converge. This in turn means that the series does not converge.

FWIW, I first tried the ratio test, which was inconclusive:

${\left\mid \frac{\frac{\sqrt{n + 1}}{3 \left(n + 1\right) + 2}}{\frac{\sqrt{n}}{3 n + 2}} \right\mid}_{n \to \infty}$

$= {\left\mid \frac{\frac{\sqrt{n} \sqrt{1 + \frac{1}{n}}}{3 n + 5}}{\frac{\sqrt{n}}{3 n + 2}} \right\mid}_{n \to \infty}$

$= {\left\mid \frac{\left(3 n + 2\right) \sqrt{1 + \frac{1}{n}}}{3 n + 5} \right\mid}_{n \to \infty}$

Dividing top and bottom by $n$ and using Binomial Series on the square root:

$= {\left\mid \frac{\left(3 + \frac{2}{n}\right) \left(1 + \left(\frac{1}{2}\right) \frac{1}{n} + O \left(\frac{1}{n} ^ 2\right)\right)}{3 + \frac{5}{n}} \right\mid}_{n \to \infty}$

$= {\left\mid \frac{3 + O \left(\frac{1}{n}\right)}{3 + O \left(\frac{1}{n}\right)} \right\mid}_{n \to \infty} = 1$

Mar 1, 2017

the series diverges

#### Explanation:

We can apply d'Alembert's ratio test:

Suppose that;

$S = {\sum}_{r = 1}^{\infty} {a}_{n} \setminus \setminus$, and $\setminus \setminus L = {\lim}_{n \rightarrow \infty} | {a}_{n + 1} / {a}_{n} |$

Then

if L < 1 then the series converges absolutely;
if L > 1 then the series is divergent;
if L = 1 or the limit fails to exist the test is inconclusive.

So our series is;

$S = {\sum}_{n = 1}^{\infty} \frac{\sqrt{n}}{3 n + 2}$

So our test limit is:

$L = {\lim}_{n \rightarrow \infty} | \frac{\frac{\sqrt{n + 1}}{3 \left(n + 1\right) + 2}}{\frac{\sqrt{n}}{3 n + 2}} |$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} | \frac{\sqrt{n + 1}}{3 \left(n + 1\right) + 2} \cdot \frac{3 n + 2}{\sqrt{n}} |$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} | \frac{\sqrt{n + 1}}{3 n + 3 + 2} \cdot \frac{3 n + 2}{\sqrt{n}} |$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} | \frac{\sqrt{n + 1}}{3 n + 5} \cdot \frac{3 n + 2}{\sqrt{n}} \cdot \frac{\frac{1}{\sqrt{n}}}{\frac{1}{\sqrt{n}}} |$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} | \left(\frac{\sqrt{n + 1}}{\sqrt{n}}\right) \cdot \frac{3 n + 2}{3 n + 5} |$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} | \left(\sqrt{\frac{n + 1}{n}}\right) \cdot \frac{3 n + 2}{3 n + 5} \cdot \frac{\frac{1}{n}}{\frac{1}{n}} |$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} | \sqrt{1 + \frac{1}{n}} \cdot \frac{3 + \frac{2}{n}}{3 + \frac{5}{n}} |$
$\setminus \setminus \setminus = | \sqrt{1 + 0} \cdot \frac{3 + 0}{3 + 0} |$
$\setminus \setminus \setminus = | 1 |$
$\setminus \setminus \setminus = 1$

And so the ratio test is inconclusive.

You can draw your own interpretation from the failure of this test, but in most cases it will be that the series does not converge..

We must therefore choose another test such as the integral test, or
We can examine the behaviour numerically. Doing so we can conclude (although not necessarily prove) that the series diverges!