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.
