# Convergence of $\sum_{n=3}^{\infty}\frac{1}{n\log n(\log\log n)^\alpha}$

Does the following series converge: $\displaystyle\sum_{n=3}^{\infty}\frac{1}{n\log n(\log\log n)^\alpha}$ and $\alpha>0$ ?

Using Cauchy condensation test twice:

\begin{align} \displaystyle\sum_{n=3}^{\infty}\frac{3^n}{3^n\log 3^n(\log\log 3^n)^\alpha} &= \displaystyle\sum_{n=3}^{\infty}\frac{1}{n\log 3(\log(n\log 3)^\alpha} \\ &= \displaystyle\sum_{n=3}^{\infty}\frac{3^n}{3^n\log 3(\log(3^n\log 3)^\alpha} \\ &= \displaystyle\sum_{n=3}^{\infty}\frac{1}{\log (3) (n\log(3\log 3)^\alpha}\end {align}

So the series converges iff $\alpha >1$ and diverges otherwise like the harmonic series.

• There's a parenthesis missing. What does the $\alpha$ exponentiate? – Daniel Fischer Apr 22 '14 at 13:14
• @DanielFischer fixed. – GinKin Apr 22 '14 at 13:16
• That works. Integral test works even faster. – André Nicolas Apr 22 '14 at 13:17
• @AndréNicolas I can't use it... – GinKin Apr 22 '14 at 13:18
• That works, but if you write it down for real, do not equate the condensed series with the uncondensed one as you did in the second equality sign. – LutzL Apr 22 '14 at 15:20