How do you find the sum of the infinite geometric series 7/8 + 7/12 + 7/18 + 7/27 ...?

Jan 3, 2016

This sum exists and is equal to $\frac{21}{8}$.

Explanation:

General form of geometric series:
$a + a q + a {q}^{2} + a {q}^{3} + \ldots$

Factor out first term $a$ to see what the quotient $q$ of this geometric series could be:

$\frac{7}{8} \left(1 + \frac{8}{7} \cdot \frac{7}{12} + \frac{8}{7} \cdot \frac{7}{18} + \frac{8}{7} \cdot \frac{7}{27} + \ldots\right) = \frac{7}{8} \left(1 + \frac{2}{3} + \frac{4}{9} + \frac{8}{27} + \ldots\right) = \frac{7}{8} \left(1 + \frac{2}{3} + {\left(\frac{2}{3}\right)}^{2} + {\left(\frac{2}{3}\right)}^{3} + \ldots\right)$

Now we could see that $a = \frac{7}{8}$ and $q = \frac{2}{3}$. As long as $- 1 < q < 1$ the series converges and sum exists:
$a + a q + a {q}^{2} + a {q}^{3} + \ldots = \frac{a}{1 - q} = \frac{\frac{7}{8}}{1 - \frac{2}{3}} = \frac{7}{8} \cdot 3 = \frac{21}{8}$