# How do you apply the ratio test to determine if sum_(n=2)^oo 10^n/(lnn)^n is convergent to divergent?

Dec 31, 2017

The Ratio Test can be used to show that this series converges.

#### Explanation:

Let ${a}_{n} = {10}^{n} / \left({\left(\ln \left(n\right)\right)}^{n}\right)$. Then

${a}_{n + 1} / {a}_{n} = {10}^{n + 1} / \left({\left(\ln \left(n + 1\right)\right)}^{n + 1}\right) \cdot \frac{{\left(\ln \left(n\right)\right)}^{n}}{10} ^ \left\{n\right\}$

$= \frac{10}{\ln} \left(n + 1\right) \cdot {\left(\ln \frac{n}{\ln} \left(n + 1\right)\right)}^{n}$

Note that ${\left(\ln \frac{n}{\ln} \left(n + 1\right)\right)}^{n} \le 1$ for all $n \ge q 2$ (L'Hopital's Rule can be used to help show that the limit of this expression is 1, but it is enough to note that it is bounded).

All this implies that

${a}_{n + 1} / {a}_{n} = \frac{10}{\ln} \left(n + 1\right) \cdot {\left(\ln \frac{n}{\ln} \left(n + 1\right)\right)}^{n} \to 0 = L < 1$ as $n \to \infty$.

Therefore, the Ratio Test implies that ${\sum}_{n = 2}^{\infty} {a}_{n} = {\sum}_{n = 2}^{\infty} {10}^{n} / \left({\left(\ln \left(n\right)\right)}^{n}\right)$ converges.