# How do you test for convergence of Sigma 5/(6n^2+n-1) from n=[1,oo)?

Jan 11, 2017

${\sum}_{n = 1}^{\infty} \frac{5}{6 {n}^{2} + n - 1}$ is convergent.

#### Explanation:

As the terms of the series are positive we can use the integral test, using:

$f \left(x\right) = \frac{5}{6 {x}^{2} + x - 1}$

We have that:

$f \left(x\right) > 0$ for x in [1,+oo)]

$f ' \left(x\right) = - \frac{5 \left(12 x + 1\right)}{{\left(6 {x}^{2} + x - 1\right)}^{2}} < 0$ for $x \in \left[1 , + \infty\right]$, so the function is monotone decreasing in that interval.

${\lim}_{x \to \infty} \frac{5}{6 {x}^{2} + x - 1} = 0$

$f \left(n\right) = \frac{5}{6 {n}^{2} + n - 1}$

so all the hypotheses of the integral test are satisfied and we can calculate:

${\int}_{1}^{\infty} \frac{5 \mathrm{dx}}{6 {x}^{2} + x - 1}$

using partial fractions:

$6 {x}^{2} + x - 1 = \left(2 x + 1\right) \left(3 x - 1\right)$

$\frac{A}{2 x + 1} + \frac{B}{3 x - 1} = \frac{5}{\left(2 x + 1\right) \left(3 x - 1\right)}$

$A \left(3 x - 1\right) + B \left(2 x + 1\right) = 5$

$\left(3 A + 2 B\right) {x}^{2} - \left(A - B\right) = 5$

{ color(white) [ color(black) ((3A+2B = 0) ,( A-B = -5) ) color(white) ] color(black)

{ color(white) [ color(black) ((A=-2 ),( B=3) ) color(white) ] color(black)

So:

${\int}_{1}^{\infty} \frac{5 \mathrm{dx}}{6 {x}^{2} + x - 1} = {\int}_{1}^{\infty} \frac{- 2 \mathrm{dx}}{2 x + 1} + {\int}_{1}^{\infty} \frac{3 \mathrm{dx}}{3 x - 1} = {\left[\ln \left(3 x - 1\right) - \ln \left(2 x + 1\right)\right]}_{1}^{\infty}$

Using the properties of logarithms:

$\ln \left(3 x - 1\right) - \ln \left(2 x + 1\right) = \ln \left(\frac{3 x - 1}{2 x + 1}\right)$

so that:

${\left[\ln \left(3 x - 1\right) - \ln \left(2 x + 1\right)\right]}_{1}^{\infty} = {\left[\ln \left(\frac{3 x - 1}{2 x + 1}\right)\right]}_{1}^{\infty}$

And as:
${\lim}_{x \to \infty} \ln \left(\frac{3 x - 1}{2 x + 1}\right) = \ln \left(\frac{3}{2}\right)$

${\left[\ln \left(\frac{3 x - 1}{2 x + 1}\right)\right]}_{1}^{\infty} = \ln \left(\frac{3}{2}\right) - \ln \left(\frac{2}{3}\right) = 2 \ln \left(\frac{3}{2}\right)$

Finally we have:

${\int}_{1}^{\infty} \frac{5 \mathrm{dx}}{6 {x}^{2} + x - 1} = 2 \ln \left(\frac{3}{2}\right)$

and as the integral is convergent, so is the series.