# How do you use the direct comparison test to determine if Sigma 3^n/(4^n+5) from [0,oo) is convergent or divergent?

Feb 1, 2018

See explanation.

#### Explanation:

${4}^{n} + 5 > {4}^{n}$ which means that $\frac{1}{{4}^{n} + 5} < \frac{1}{4} ^ n$.

Since ${3}^{n} > 0$, ${3}^{n} / \left({4}^{n} + 5\right) < {3}^{n} / {4}^{n}$ for all $n \ge 0$.

Given this, we know $\sum {3}^{n} / \left({4}^{n} + 5\right) < \sum {3}^{n} / {4}^{n}$.

$\sum {\left(\frac{3}{4}\right)}^{n}$ is a convergent geometric series.

Since $\sum {3}^{n} / \left({4}^{n} + 5\right)$ is positive termed and $\sum {3}^{n} / \left({4}^{n} + 5\right) < \sum {3}^{n} / {4}^{n}$, $\sum {3}^{n} / \left({4}^{n} + 5\right)$ must converge by direct comparison.