# How do you test the series Sigma (5^n+6^n)/(2^n+7^n) from n is [0,oo) for convergence?

Mar 20, 2017

The series:

${\sum}_{n = 0}^{\infty} \left(\frac{{5}^{n} + {6}^{n}}{{2}^{n} + {7}^{n}}\right)$

is convergent.

#### Explanation:

Use the ratio test by evaluating:

$\left\mid {a}_{n + 1} / {a}_{n} \right\mid = \left(\frac{\frac{{5}^{n + 1} + {6}^{n + 1}}{{2}^{n + 1} + {7}^{n + 1}}}{\frac{{5}^{n} + {6}^{n}}{{2}^{n} + {7}^{n}}}\right)$

$\left\mid {a}_{n + 1} / {a}_{n} \right\mid = \left(\frac{{5}^{n + 1} + {6}^{n + 1}}{{2}^{n + 1} + {7}^{n + 1}}\right) \left(\frac{{2}^{n} + {7}^{n}}{{5}^{n} + {6}^{n}}\right)$

$\left\mid {a}_{n + 1} / {a}_{n} \right\mid = \left(\frac{{5}^{n + 1} + {6}^{n + 1}}{{5}^{n} + {6}^{n}}\right) \left(\frac{{2}^{n} + {7}^{n}}{{2}^{n + 1} + {7}^{n + 1}}\right)$

$\left\mid {a}_{n + 1} / {a}_{n} \right\mid = \left({6}^{n + 1} / {6}^{n}\right) \left(\frac{{\left(\frac{5}{6}\right)}^{n + 1} + 1}{{\left(\frac{5}{6}\right)}^{n} + 1}\right) \left({7}^{n} / {7}^{n + 1}\right) \left(\frac{{\left(\frac{2}{7}\right)}^{n} + 1}{{\left(\frac{2}{7}\right)}^{n + 1} + 1}\right)$

$\left\mid {a}_{n + 1} / {a}_{n} \right\mid = \left(\frac{6}{7}\right) \left(\frac{{\left(\frac{5}{6}\right)}^{n + 1} + 1}{{\left(\frac{5}{6}\right)}^{n} + 1}\right) \left(\frac{{\left(\frac{2}{7}\right)}^{n} + 1}{{\left(\frac{2}{7}\right)}^{n + 1} + 1}\right)$

We can now pass to the limit for $n \to \infty$:

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = {\lim}_{n \to \infty} \left(\frac{6}{7}\right) \left(\frac{{\left(\frac{5}{6}\right)}^{n + 1} + 1}{{\left(\frac{5}{6}\right)}^{n} + 1}\right) \left(\frac{{\left(\frac{2}{7}\right)}^{n} + 1}{{\left(\frac{2}{7}\right)}^{n + 1} + 1}\right)$

Now as $\frac{5}{6} < 1$ we have:

${\lim}_{n \to \infty} {\left(\frac{5}{6}\right)}^{n} = {\lim}_{n \to \infty} {\left(\frac{5}{6}\right)}^{n + 1} = 0$

and similarly:

${\lim}_{n \to \infty} {\left(\frac{2}{7}\right)}^{n} = {\lim}_{n \to \infty} {\left(\frac{2}{7}\right)}^{n + 1} = 0$

So:

${\lim}_{n \to \infty} \left(\frac{6}{7}\right) \left(\frac{{\left(\frac{5}{6}\right)}^{n + 1} + 1}{{\left(\frac{5}{6}\right)}^{n} + 1}\right) \left(\frac{{\left(\frac{2}{7}\right)}^{n} + 1}{{\left(\frac{2}{7}\right)}^{n + 1} + 1}\right) = \frac{6}{7} < 1$

which proves the series to be convergent.