# Test for convergence of the series $\sum_{n=2}^\infty\frac{1}{(\ln n)^{\ln n}}$

Could I have a hint for testing the convergence of the following series please?

$$\sum_{n=2}^\infty\frac{1}{(\ln n)^{\ln n}}$$

Edit

The integral test does not work because $\int_1^n\frac{1}{(\ln x)^{\ln x}}dx$ has not an elementary primitive.

Thank You.

• did you try the integral test? – Alex Jan 26 '14 at 19:57
• en.wikipedia.org/wiki/Sophomore's_dream – Alex Jan 26 '14 at 20:00
• @Alex I'm trying with the integral test, but this integral is really tough for me... I'll check your link now! Thank you for your help man; I really appreciate it. – Charlie Jan 26 '14 at 20:03
• @Alex Am I on the right way? Can this integral be reduced to a simple one? – Charlie Jan 26 '14 at 20:41
• – Hans Lundmark Dec 9 '17 at 14:45

Alternate hint:

$$(\ln n)^{\ln n} = n^{\ln \ln n}.$$

• Thank you. I did it with the comparison test. – Charlie Jan 26 '14 at 22:18
• You're welcome :) – Antonio Vargas Jan 26 '14 at 23:31
• I have never seen this before, how do you get to that equation? – Nhat Jan 27 '14 at 17:20
• @KitKat, take $\ln$ of $(\ln n)^{\ln n}$ to get $(\ln n)\ln\ln n$, rewrite this as $\ln(n^{\ln\ln n})$, then exponentiate this to get $n^{\ln \ln n}$. – Antonio Vargas Jan 27 '14 at 17:38
• @AntonioVargas Nice, thanks. – Nhat Jan 27 '14 at 19:52

Hint:

Consider

$$2^k \frac{1}{(k\ln e)^{k\ln 2}}>\sum_{2^k\le n< 2^{k+1}} \frac{1}{(\ln n)^{\ln n}}> 2^k\frac{1}{((k+1)\ln e)^{(k+1)\ln 2}},$$