# How do you find the sum of the infinite geometric series (3/2)-(5/6)+(7/18)-...?

Jun 28, 2016

$\frac{15}{16}$

#### Explanation:

This is not a geometric series. It has no common ratio, but the intention seems to be that the general term is:

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

If we sum pairs of terms, then the resulting series has all positive terms, with the following expression for the general term:

${b}_{n} = \frac{\left(4 n - 2\right)}{\left(3 \cdot {9}^{n - 1}\right)}$

The numerators are in arithmetic progression with initial term $2$ and common difference $4$.

We can split this series into a series of series, hence into a product of a geometric series and a geometric series with constant offset.

Note first that ${\sum}_{k = 0}^{\infty} \frac{1}{9} ^ k = \frac{9}{8}$

So we find:

$\frac{3}{2} - \frac{5}{6} + \frac{7}{18} - \frac{9}{54} + \frac{11}{162} - \frac{13}{486} + \frac{15}{1458} - \frac{17}{4374} + \ldots$

$= \frac{2}{3} + \frac{6}{27} + \frac{10}{243} + \frac{14}{2187} + \ldots$

$= \frac{1}{3} \left(\left(2 + \frac{2}{9} + \frac{2}{9} ^ 2 + \frac{2}{9} ^ 3 + \ldots\right) + \left(\frac{4}{9} + \frac{4}{9} ^ 2 + \frac{4}{9} ^ 3 + \ldots\right) + \left(\frac{4}{9} ^ 2 + \frac{4}{9} ^ 3 + \frac{4}{9} ^ 4 + \ldots\right) + \left(\frac{4}{9} ^ 3 + \frac{4}{9} ^ 4 + \frac{4}{9} ^ 5 + \ldots\right) + \ldots\right)$

$= \frac{1}{3} \left(1 + \frac{1}{9} + \frac{1}{9} ^ 2 + \frac{1}{9} ^ 3 + \ldots\right) \left(2 + \frac{4}{9} + \frac{4}{9} ^ 2 + \frac{4}{9} ^ 3 + \frac{4}{9} ^ 4 + \ldots\right)$

$= \frac{4}{3} \left({\sum}_{k = 0}^{\infty} \frac{1}{9} ^ k\right) \left(- \frac{1}{2} + {\sum}_{k = 0}^{\infty} \frac{1}{9} ^ k\right)$

$= \frac{4}{3} \cdot \frac{9}{8} \cdot \frac{9 - 4}{8}$

$= \frac{4}{3} \cdot \frac{9}{8} \cdot \frac{5}{8}$

$= \frac{15}{16}$