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

Question

Elijah Abram · Accepted Answer

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

The ratio test states that: #if lim_(n->oo) abs(a_(n+1)/a_n) < 1#, then #sum_(n=k)^oo a_n# converges for any integer #k#. #if lim_(n->oo) abs(a_(n+1)/a_n) > 1#, then #sum_(n=k)^oo a_n# diverges for any integer #k#. #if lim_(n->oo) abs(a_(n+1)/a_n) = 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: #color(white)"XXX" lim_(n->oo) abs( (1/(ln(ln(n+1))^(n+1))) / (1/(ln(ln(n))^n) )) = lim_(n->oo) abs( (ln(ln(n))^n) / (ln(ln(n+1))^(n+1)) )# # 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^oo#. All is not lost, though. This is when logic and reasoning come into play. #ln(ln(n)) / ln(ln(n+1))# is never greater than 1 for any finite positive number n. So, for all natural numbers #n#, #(ln(ln(n)) / ln(ln(n+1)))^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->oo) (ln(ln(n))/ln(ln(n+1)))^n = lim_(n->oo) N# #0 <= lim_(n->oo) N <= 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) )# #= (lim_(n->oo) N) * 0# #= 0 color(white)"XX"# (since #lim_(n->oo) N # is not infinite) The ratio approaches 0 as n approaches infinity, so #sum_(n=3)^oo 1/(ln(ln(n))^n# converges.

Evelyn Agard · Answer

undefined To apply the ratio test to determine if $\sum_{n=3}^{\infty} \frac{1}{(\ln(\ln n))^n}$ is convergent or divergent, we compute the limit:

$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$$

where $a_n = \frac{1}{(\ln(\ln n))^n}$.

Compute $a_{n+1}$ and $a_n$, then take their ratio. If the limit is less than 1, the series is convergent; if it's greater than 1 or diverges to infinity, the series is divergent; if it's exactly 1, the ratio test is inconclusive.

After taking the limit, you'll need to analyze whether it's less than, greater than, or equal to 1.