Question

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

Answer 1

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

Explanation:

Let `a_{n}=10^{n}/((ln(n))^{n})`. Then

`a_{n+1}/a_{n}=10^{n+1}/((ln(n+1))^{n+1}) * ((ln(n))^{n})/10^{n}`

`=10/ln(n+1) * (ln(n)/ln(n+1))^{n}`

Note that `(ln(n)/ln(n+1))^{n} <= 1` for all `n geq 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}=10/ln(n+1) * (ln(n)/ln(n+1))^{n}->0=L < 1` as `n->infty`.

Therefore, the Ratio Test implies that `sum_{n=2}^{infty}a_{n}=sum_{n=2}^{infty}10^{n}/((ln(n))^{n})` converges.