Series

**RU2010** · September 13th 2010, 07:12 PM

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?

**pickslides** · September 13th 2010, 07:53 PM

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

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

**acevipa** · September 13th 2010, 08:38 PM

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

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

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

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

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

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

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

$=\dfrac{1}{8}$

**acevipa** · September 13th 2010, 08:39 PM

EDIT: Sorry double post

**RU2010** · September 13th 2010, 09:16 PM

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

**CaptainBlack** · September 13th 2010, 11:22 PM

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

Thread: Series

LinkBack

Thread Tools

Display

Series

Similar Math Help Forum Discussions

Show that a series converges absolutely given convergence of two related series

Exploiting geometric series with power series to find Taylors series

question about power series, taylor series, maclaurin series

calculus II areas about axis, pressure, infinite series, MacLaurin series

Series and Sequence/Application (sum of a geometric series) - Supperannuation

Search Tags