# How do you determine if Sigma (7*5^n)/6^n from n=[0,oo) converge and find the sums when they exist?

Feb 8, 2017

${\sum}_{n = 0}^{\infty} \frac{7 \cdot {5}^{n}}{6} ^ n = 42$

#### Explanation:

Write the series as:

${\sum}_{n = 0}^{\infty} \frac{7 \cdot {5}^{n}}{6} ^ n = 7 {\sum}_{n = 0}^{\infty} {\left(\frac{5}{6}\right)}^{n}$

so this is a geometric series of ratio $r = \frac{5}{6}$.

As $\left\mid r \right\mid < 1$ the series is convergent and its sum is:

$7 {\sum}_{n = 0}^{\infty} {\left(\frac{5}{6}\right)}^{n} = \frac{7}{1 - \frac{5}{6}} = \frac{7}{\frac{1}{6}} = 42$