# Math Help - Series

1. ## Series

How would I calculate the sum of the following series:

$\sum _{n=1}^{\infty }{\frac { \left( -2 \right) ^{n-1}}{{6}^{n}}}$

Of course I can use the calculator, but how would I do it manually?

2. Have you tried to simplify the summation?

$\frac{(-2)^{n-1}}{6^n}= \frac{(-2)^{n}(-2)^{-1}}{(-3\times -2)^n} = \frac{(-2)^{n}(-2)^{-1}}{(-3)^n\times (-2)^n} = \frac{(-2)^{-1}}{(-3)^n}$

or even noticed the first couple of terms

$\frac{1}{6}-\frac{1}{18}+\frac{1}{54}-\frac{1}{162}+\dots$

3. $\sum _{n=1}^{\infty }{\frac { \left( -2 \right) ^{n-1}}{{6}^{n}}}=\sum _{n=1}^{\infty }{\frac { \left( -2 \right) ^{n-1}}{{6}^{n-1}6}}$

$=\sum _{n=1}^{\infty }{\frac{1}{6}\left(\frac{-2}{6}\right)^{n-1}}$

$=\sum _{n=1}^{\infty }{\frac{1}{6}\left(\frac{-1}{3}\right)^{n-1}}$

This is a geometric series with $a=\frac{1}{6}, r=-\frac{1}{3}$

$=\dfrac{\frac{1}{6}}{1-\left(-\frac{1}{3}\right)}$

$=\dfrac{\frac{1}{6}}{1+\frac{1}{3}}$

$=\dfrac{\frac{1}{6}}{\frac{4}{3}}$

$=\dfrac{1}{8}$

4. EDIT: Sorry double post

5. Thanks so much! This makes perfect sense once you showed how this is a geometric series!

6. Originally Posted by RU2010
How would I calculate the sum of the following series:

$\sum _{n=1}^{\infty }{\frac { \left( -2 \right) ^{n-1}}{{6}^{n}}}$

Of course I can use the calculator, but how would I do it manually?
$\displaystyle \sum _{n=1}^{\infty }\frac { ( -2 ) ^{n-1}}{6^{n}}=-\frac{1}{2} \sum _{n=1}^{\infty }\left(-\;\frac { 1 }{3}\right)^n$

and the sum on the right is a geometric series.

CB