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.