Test whether the series is convergent or divergent. The summation of 1/(ln n)^ln n where n=1 goes 2 infinity.?

Question

∑_n=1 , ∞; 1/(ln n)^ln n. Is it convergent or divergent? Here's a soution from my friend, (ln n)^ln n = [e^ln ln n]^ln n = (e^ln n) ^lnlnn and ln ln n -> ∞ as n-> ∞, so ln ln n >2 for sufficiently large n. For these the n we have (ln n)^ln n > n^2 so taking the reciprocal n^2 would be larger. Since 1/n^2 converges by p-series so does 1/(ln n)^ln n by the comparison test. I can't seem to understand what the relationship is and can someone explain how to get this (ln n)^ln n > n^2 ? Thanks in advance!

Answer 1

Remember, you are talking about limits as n gets big. This identity is probably not true for all values of n (it isn't - it fails for 1), but you don't care.

Choose m so that n = e^m (m is not an integer). Then
ln(n) = m, ln(n)^ln(n) = m^m, and n^2 = e^2m. When is m^m > e^2m? If m > e^2 this identity holds, so for m > 27 (= 3^3 > e^e) it is true.

Answer 2

ln^ln =e^ln*lln = (e^ln)^lln. But e^ln is n So ln^ln = n^lln.
As lln => to infinity lln>2 from n> e^(e^2) and 1/n^lln <1/n^2

Answer 3

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