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.

  • $\begingroup$ There's a parenthesis missing. What does the $\alpha$ exponentiate? $\endgroup$ – Daniel Fischer Apr 22 '14 at 13:14
  • $\begingroup$ @DanielFischer fixed. $\endgroup$ – GinKin Apr 22 '14 at 13:16
  • 2
    $\begingroup$ That works. Integral test works even faster. $\endgroup$ – André Nicolas Apr 22 '14 at 13:17
  • $\begingroup$ @AndréNicolas I can't use it... $\endgroup$ – GinKin Apr 22 '14 at 13:18
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    $\begingroup$ 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. $\endgroup$ – LutzL Apr 22 '14 at 15:20

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