# Show that ∑1/(log n)^(log n) converges

## Homework Statement

Express (log n)log n as a power of n, and use the result to show that ∑1/(log n)log n converges.

blah blah

## The Attempt at a Solution

np(n) = (log n)log n
----> log[ np(n) ] = log[ (log n)log n ]
----> p(n) (log n) = (log n) log(log n)
----> p(n) = log(log n)

----> ∑1/(log n)(log n) = ∑ 1/nlog(log n)np(n) = (log n)log n
----> log[ np(n) ] = log[ (log n)log n ]
----> p(n) (log n) = (log n) log(log n)
----> p(n) = log(log n)

----> ∑1/(log n)(log n) = ∑ 1/nlog(log n)

.... Now what? I can't think of any comparison test or anything for un = 1/nlog(log n)

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Try taking n greater than e raised to the power e.

Try taking n greater than e raised to the power e.
Huh? Explain a tad bit more.

The series with terms 1/n^p converges whenever p is greater than one. When n is greater than e raised to the power e, log (log n) is greater than 1. There are only finitely many terms of your sequence for which n is less than e raised to the power e; apply the comparison test accordingly.

The series with terms 1/n^p converges whenever p is greater than one. When n is greater than e raised to the power e, log (log n) is greater than 1. There are only finitely many terms of your sequence for which n is less than e raised to the power e; apply the comparison test accordingly.
Ah, I see!

n > ee ---> log(n) > log(ee) --> log(n) > e log(e) ---> log(n) > e ---> log(log(n)) > log(e) ----> log(log(n)) > 1.

Let 1 < p < log(log(n)). Then 1/nlog(log(n)) < 1/np whenever n > ee.

Because ∑1/np converges for all p > 1,

∑1/nlog(log(n) converges as well (Theorem 19.2.II)

The only problem is that I don't think we've learned ∑1/np converges for all p > 1. I can't be spouting off stuff willy-nilly.