# How do you find the sum of the infinite geometric series (1/4)+(1/10)+(1/18)+(1/28)+(1/40)+...?

Mar 30, 2016

Derive general formula for terms, split into partial fractions, then sum and simplify to find that sum is $\frac{11}{18}$

#### Explanation:

This is not a geometric series. There is no common ratio between terms.

We can derive a general formula for a term of the series as follows:

Write down the sequence of reciprocals of the terms:

$\textcolor{b l u e}{4} , 10 , 18 , 28 , 40$

Write down the sequence of differences of that sequence:

$\textcolor{b l u e}{6} , 8 , 10 , 12$

Write down the sequence of differences of that sequence:

$\textcolor{b l u e}{2} , 2 , 2$

Having reached a constant sequence, we can write a formula for a general term of the original series using the initial term of each of these sequences as coefficients:

a_n = 1/(color(blue)(4)/(0!) + color(blue)(6)/(1!)(n-1) + color(blue)(2)/(2!)(n-1)(n-2))

$= \frac{1}{4 + 6 n - 6 + {n}^{2} - 3 n + 2}$

$= \frac{1}{{n}^{2} + 3 n}$

Next, try to express this as a partial fraction expansion as follows:

$\frac{1}{{n}^{2} + 3 n} = \frac{A}{n} + \frac{B}{n + 3}$

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

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

Equating coefficients we find:

$\left\{\begin{matrix}A + B = 0 \\ 3 A = 1\end{matrix}\right.$

Hence $A = \frac{1}{3}$ and $B = - \frac{1}{3}$

So:

${\sum}_{n = 1}^{N} \frac{1}{{n}^{2} + 3 n}$

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

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

$= \frac{1}{3} \left({\sum}_{n = 1}^{3} \frac{1}{n} + {\sum}_{n = 4}^{N} \frac{1}{n} - {\sum}_{n = 1}^{N - 3} \frac{1}{n + 3} - {\sum}_{n = N - 2}^{N} \frac{1}{n + 3}\right)$

$= \frac{1}{3} \left(\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 4}^{N} \frac{1}{n}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 4}^{N} \frac{1}{n}}}} - \frac{1}{N + 1} - \frac{1}{N + 2} - \frac{1}{N + 3}\right)$

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

$= \frac{1}{3} \left(\frac{11}{6} - \frac{1}{N + 1} - \frac{1}{N + 2} - \frac{1}{N + 3}\right)$

So:

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

$= {\lim}_{N \to \infty} {\sum}_{n = 1}^{N} \frac{1}{{n}^{2} + 3 n}$

$= {\lim}_{N \to \infty} \left(\frac{1}{3} \left(\frac{11}{6} - \frac{1}{N + 1} - \frac{1}{N + 2} - \frac{1}{N + 3}\right)\right)$

$= \frac{1}{3} \cdot \frac{11}{6}$

$= \frac{11}{18}$