How would I calculate the sum of the following series:
$\displaystyle \sum _{n=1}^{\infty }{\frac { \left( -2 \right) ^{n-1}}{{6}^{n}}}$
Of course I can use the calculator, but how would I do it manually?
Have you tried to simplify the summation?
$\displaystyle \frac{(-2)^{n-1}}{6^n}= \frac{(-2)^{n}(-2)^{-1}}{(-3\times -2)^n} = \frac{(-2)^{n}(-2)^{-1}}{(-3)^n\times (-2)^n} = \frac{(-2)^{-1}}{(-3)^n}$
or even noticed the first couple of terms
$\displaystyle \frac{1}{6}-\frac{1}{18}+\frac{1}{54}-\frac{1}{162}+\dots$
$\displaystyle \sum _{n=1}^{\infty }{\frac { \left( -2 \right) ^{n-1}}{{6}^{n}}}=\sum _{n=1}^{\infty }{\frac { \left( -2 \right) ^{n-1}}{{6}^{n-1}6}}$
$\displaystyle =\sum _{n=1}^{\infty }{\frac{1}{6}\left(\frac{-2}{6}\right)^{n-1}}$
$\displaystyle =\sum _{n=1}^{\infty }{\frac{1}{6}\left(\frac{-1}{3}\right)^{n-1}}$
This is a geometric series with $\displaystyle a=\frac{1}{6}, r=-\frac{1}{3}$
$\displaystyle =\dfrac{\frac{1}{6}}{1-\left(-\frac{1}{3}\right)}$
$\displaystyle =\dfrac{\frac{1}{6}}{1+\frac{1}{3}}$
$\displaystyle =\dfrac{\frac{1}{6}}{\frac{4}{3}}$
$\displaystyle =\dfrac{1}{8}$