# How do you test the series Sigma 5^n/(3^n+4^n) from n is [0,oo) for convergence?

Jan 17, 2017

The series is divergent.

#### Explanation:

${5}^{n} / \left({3}^{n} + {4}^{n}\right) = \frac{1}{{\left(\frac{3}{5}\right)}^{2} + {\left(\frac{4}{5}\right)}^{n}}$ but

$\frac{1}{2 {\left(\frac{4}{5}\right)}^{2}} < \frac{1}{{\left(\frac{3}{5}\right)}^{2} + {\left(\frac{4}{5}\right)}^{n}} < \frac{1}{2 {\left(\frac{3}{5}\right)}^{2}}$ or

$\frac{1}{2} {\left(\frac{5}{4}\right)}^{n} < \frac{1}{{\left(\frac{3}{5}\right)}^{2} + {\left(\frac{4}{5}\right)}^{n}} < \frac{1}{2} {\left(\frac{5}{3}\right)}^{n}$ but

$\frac{5}{4} > 1$ and $\frac{5}{3} > 1$ so

${\sum}_{n = 1}^{\infty} {\left(\frac{5}{4}\right)}^{n} = \infty$ and ${\sum}_{n = 1}^{\infty} {\left(\frac{5}{3}\right)}^{n} = \infty$

so

${\sum}_{n = 0}^{\infty} {5}^{n} / \left({3}^{n} + {4}^{n}\right) = \infty$ being divergent
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