# How to solve sum_(n=1)^oo 1/((3n-2)(3n+1)) using partial fractions?

## ${\sum}_{n = 1}^{\infty} \frac{1}{\left(3 n - 2\right) \left(3 n + 1\right)}$ There is an infinity sign on the top of the summation sign and n=1 on the bottom. Please write in details; I don't get how the answer is $\frac{1}{12}$.

##### 1 Answer
Nov 25, 2017

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

#### Explanation:

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

So:

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

Putting $n = \frac{2}{3}$ this becomes:

$1 = 3 A \text{ }$ so $\text{ } A = \frac{1}{3}$

Putting $n = - \frac{1}{3}$ it becomes:

$1 = - 3 B \text{ }$ so $\text{ } B = - \frac{1}{3}$

Then:

${\sum}_{n = 1}^{N} \frac{1}{\left(3 n - 2\right) \left(3 n + 1\right)} = {\sum}_{n = 1}^{N} \left(\frac{1}{3 \left(3 n - 2\right)} - \frac{1}{3 \left(3 n + 1\right)}\right)$

$\textcolor{w h i t e}{{\sum}_{n = 1}^{N} \frac{1}{\left(3 n - 2\right) \left(3 n + 1\right)}} = {\sum}_{n = 1}^{N} \frac{1}{3 \left(3 n - 2\right)} - {\sum}_{n = 1}^{N} \frac{1}{3 \left(3 n + 1\right)}$

$\textcolor{w h i t e}{{\sum}_{n = 1}^{N} \frac{1}{\left(3 n - 2\right) \left(3 n + 1\right)}} = {\sum}_{n = 1}^{N} \frac{1}{3 \left(3 n - 2\right)} - {\sum}_{n = 2}^{N + 1} \frac{1}{3 \left(3 n - 2\right)}$

color(white)(sum_(n=1)^N 1/((3n-2)(3n+1))) = 1/3+color(red)(cancel(color(black)(sum_(n=2)^N 1/(3(3n-2))))) - color(red)(cancel(color(black)(sum_(n=2)^N 1/(3(3n-2))))) - 1/(3(3N+1)

color(white)(sum_(n=1)^N 1/((3n-2)(3n+1))) = 1/3 - 1/(3(3N+1)

So:

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

$\textcolor{w h i t e}{{\sum}_{n = 1}^{\infty} \frac{1}{\left(3 n - 2\right) \left(3 n + 1\right)}} = {\lim}_{N \to \infty} \left(\frac{1}{3} - \frac{1}{3 \left(3 N + 1\right)}\right) = \frac{1}{3}$

Note that:

${\sum}_{n = \textcolor{b l u e}{2}}^{\infty} \frac{1}{\left(3 n - 2\right) \left(3 n + 1\right)} = \frac{1}{3 \left(3 \left(\textcolor{b l u e}{2}\right) - 2\right)} = \frac{1}{12}$