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

[Jamin2112]

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.

Homework Equations

blah blah

The Attempt at a Solution

n^p(n) = (log n)^{log n}
----> log[ n^p(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/n^{log(log n)}n^p(n) = (log n)^{log n}
----> log[ n^p(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/n^{log(log n)}

... Now what? I can't think of any comparison test or anything for u_n = 1/n^{log(log n)}

 

[ns2675]

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

 

[Jamin2112]

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

Huh? Explain a tad bit more.

 

[ns2675]

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.

 

[Jamin2112]

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 > e^e ---> log(n) > log(e^e) --> 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/n^log(log(n)) < 1/n^p whenever n > e^e.

Because ∑1/n^p converges for all p > 1,

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

 

[Jamin2112]

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