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

Mar 1, 2017

See below.

#### Explanation:

$\frac{{7}^{n} - {6}^{n}}{5} ^ n = {\left(\frac{7}{5}\right)}^{n} - {\left(\frac{6}{5}\right)}^{n}$

$\frac{{\left(\frac{7}{5}\right)}^{n} - {\left(\frac{6}{5}\right)}^{n}}{\frac{7}{5} - \frac{6}{5}} = {\left(\frac{7}{5}\right)}^{n - 1} + {\left(\frac{7}{5}\right)}^{n - 2} \left(\frac{6}{5}\right) + \cdots + \left(\frac{7}{5}\right) {\left(\frac{6}{5}\right)}^{n - 2} + {\left(\frac{6}{5}\right)}^{n - 1}$ or

${\left(\frac{7}{5}\right)}^{n} - {\left(\frac{6}{5}\right)}^{n} = \frac{1}{5} \left({\left(\frac{7}{5}\right)}^{n - 1} + {\left(\frac{7}{5}\right)}^{n - 2} \left(\frac{6}{5}\right) + \cdots + \left(\frac{7}{5}\right)\right)$

Here $\frac{7}{5} > 1$ and $\frac{6}{5} > 1$ so for $n > 5$ we surely will have

$\frac{1}{5} \left({\left(\frac{7}{5}\right)}^{n - 1} + {\left(\frac{7}{5}\right)}^{n - 2} \left(\frac{6}{5}\right) + \cdots + \left(\frac{7}{5}\right)\right) > 1$

Concluding this is a divergent series.