# Is the series \sum_(n=1)^\infty((-5)^(2n))/(n^2 9^n) absolutely convergent, conditionally convergent or divergent?

## (Use the appropriate test)

Apr 20, 2018

Diverges by the Ratio Test.

#### Explanation:

We'll check for Absolute Convergence using the Ratio Test, where ${a}_{n} = {\left(- 5\right)}^{2 n} / \left({n}^{2} {9}^{n}\right)$.

The Ratio Test tells us we take

$L = {\lim}_{n \to \infty} | {a}_{n + 1} / {a}_{n} | ,$ and

$L < 1$ implies absolute convergence (and hence convergence),

$L > 1$ implies divergence,

$L = 1$ is inconclusive and we need to use another test.

Then,

${a}_{n + 1} = {\left(- 5\right)}^{2 \left(n + 1\right)} / \left({\left(n + 1\right)}^{2} {9}^{n + 1}\right) = {\left(- 5\right)}^{2 n + 2} / \left({\left(n + 1\right)}^{2} {9}^{n + 1}\right)$

Thus,

$L = {\lim}_{n \to \infty} | {\left(- 5\right)}^{2 n + 2} / \left({\left(n + 1\right)}^{2} {9}^{n + 1}\right) \cdot \frac{{n}^{2} {9}^{n}}{- 5} ^ \left(2 n\right) |$

${\left(- 5\right)}^{2 n + 2} / {\left(- 5\right)}^{2 n} = {\left(- 5\right)}^{2} = 25$

${9}^{n} / {9}^{n + 1} = \frac{1}{9}$

We can factor these constants outside of the limit, getting

$\frac{25}{9} {\lim}_{n \to \infty} {n}^{2} / {\left(n + 1\right)}^{2} = \frac{25}{9} > 1$

We dropped the absolute value bars as everything is positive when we go to infinity.

We also then have divergence by the Ratio Test.