# Find the sum of the series 1/2-1/3+1/4-1/9+1/8-1/27+........ to infinity?

Jan 16, 2017

$\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{9} + \frac{1}{8} - \frac{1}{27} + \ldots \ldots . . = \frac{1}{2}$

#### Explanation:

$\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{9} + \frac{1}{8} - \frac{1}{27} + \ldots \ldots . .$ can be rewritten as

$\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \ldots . .\right) - \left(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots \ldots . .\right)$

Hence, the given infinitive is one infinitive geometric series subtracted from another infinitive geometric series.

Observe that in the infinite series $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \ldots . .$, while first term $\frac{1}{2}$ and common ratio is $\frac{1}{2}$

and in the infinite series $\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots \ldots . .$, first term $\frac{1}{3}$ and common ratio is $\frac{1}{3}$

As for an infinite series whose first term is $a$ and common ratio $r$ is such that $| r | < 1$, the sum converges to $\frac{a}{1 - r}$.

Hence $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \ldots . . = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$, and

$\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots \ldots . . = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$

Hence $\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{9} + \frac{1}{8} - \frac{1}{27} + \ldots \ldots . . = 1 - \frac{1}{2} = \frac{1}{2}$