# sum_(n=0)^oo 5^n/(3^n +4^n). Does the series converge or diverge?

## I used Divergence Test and I'm stuck on this part ${\lim}_{n \to \infty} {\left(\frac{5}{4}\right)}^{n} / \left[{\left(\frac{3}{4}\right)}^{n} + 1\right]$ Im having trouble finding the limit

Apr 9, 2017

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

#### Explanation:

You are on the good path:

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

Now as:

${\lim}_{n \to \infty} \frac{1}{1 + {\left(\frac{3}{4}\right)}^{n}} = 1$

we have:

${\lim}_{n \to \infty} {\left(\frac{5}{4}\right)}^{n} / \left(1 + {\left(\frac{3}{4}\right)}^{n}\right) = {\lim}_{n \to \infty} {\left(\frac{5}{4}\right)}^{n} \cdot {\lim}_{n \to \infty} \frac{1}{1 + {\left(\frac{3}{4}\right)}^{n}} = + \infty \cdot 1 = + \infty$

and as the sequence of the terms is not infinitesimal, the series cannot converge. As the terms are all positive the series is then divergent.

Apr 13, 2017

(I know this isn't the method you requested, but this is how I first approached the problem. With series problems, there are frequently many valid solutions.)

#### Explanation:

Note that ${3}^{n} + {4}^{n} < {4}^{n} + {4}^{n} = 2 \left({4}^{n}\right)$.

Then, ${5}^{n} / \left({3}^{n} + {4}^{n}\right) > {5}^{n} / \left(2 \left({4}^{n}\right)\right)$ since the first one has a lesser denominator.

We should see that ${\sum}_{n = 0}^{\infty} {5}^{n} / \left(2 \left({4}^{n}\right)\right) = \frac{1}{2} {\sum}_{n = 0}^{\infty} {\left(\frac{5}{4}\right)}^{n}$, which is divergent by the Geometric Series test since divergent since $\left\mid \frac{5}{4} \right\mid > 1$.

Then, by the direct comparison test, ${\sum}_{n = 0}^{\infty} {5}^{n} / \left({3}^{n} + {4}^{n}\right)$ is divergent as well since it is larger than another divergent series.