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.
