# How do you apply the ratio test to determine if Sigma 1/(ln(lnn))^n from n=[3,oo) is convergent to divergent?

##### 1 Answer
Mar 19, 2017

sum_(n=3)^oo 1/(ln(ln(n))^n converges.

#### Explanation:

The ratio test states that:

$\mathmr{if} {\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid < 1$, then ${\sum}_{n = k}^{\infty} {a}_{n}$ converges for any integer $k$.

$\mathmr{if} {\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid > 1$, then ${\sum}_{n = k}^{\infty} {a}_{n}$ diverges for any integer $k$.

$\mathmr{if} {\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = 1$, then the test is inconclusive (i.e. the series could be either convergent or divergent).

Therefore, using the ratio test for the example above gives:

$\textcolor{w h i t e}{\text{XXX}} {\lim}_{n \to \infty} \left\mid \frac{\frac{1}{\ln {\left(\ln \left(n + 1\right)\right)}^{n + 1}}}{\frac{1}{\ln {\left(\ln \left(n\right)\right)}^{n}}} \right\mid = {\lim}_{n \to \infty} \left\mid \frac{\ln {\left(\ln \left(n\right)\right)}^{n}}{\ln {\left(\ln \left(n + 1\right)\right)}^{n + 1}} \right\mid$

 color(white)"X" = lim_(n->oo) abs((ln(ln(n))/ln(ln(n+1)))^n ) * lim_(n->oo) abs( 1/(ln(ln(n+1)) )

The right limit clearly goes to zero as n approaches infinity. However, the left limit is indeterminate since it will approach the indeterminate form ${1}^{\infty}$.

All is not lost, though. This is when logic and reasoning come into play. $\ln \frac{\ln \left(n\right)}{\ln} \left(\ln \left(n + 1\right)\right)$ is never greater than 1 for any finite positive number n. So, for all natural numbers $n$, ${\left(\ln \frac{\ln \left(n\right)}{\ln} \left(\ln \left(n + 1\right)\right)\right)}^{n} = N$ where N is some number between 0 and 1.

This is true since any number between 0 and 1, when raised to any positive power, must always still be between 0 and 1.

Therefore, we can say that ${\lim}_{n \to \infty} {\left(\ln \frac{\ln \left(n\right)}{\ln} \left(\ln \left(n + 1\right)\right)\right)}^{n} = {\lim}_{n \to \infty} N$

$0 \le {\lim}_{n \to \infty} N \le 1$

Back to the original problem:

color(white)"XX" lim_(n->oo) abs((ln(ln(n))/ln(ln(n+1)))^n ) * lim_(n->oo) abs( 1/(ln(ln(n+1))^(n+1) )

$= \left({\lim}_{n \to \infty} N\right) \cdot 0$

$= 0 \textcolor{w h i t e}{\text{XX}}$ (since ${\lim}_{n \to \infty} N$ is not infinite)

The ratio approaches 0 as n approaches infinity, so sum_(n=3)^oo 1/(ln(ln(n))^n converges.