What is the quickest way to show when

$$ S(\alpha_1,\alpha_2,\cdots,\alpha_k) = \sum\limits_{n=3}^\infty \frac{1}{n (\log n)^{\alpha_1}\cdots (\log^k(n))^{\alpha_k}} $$

converges, where $\log^k(n)$ is the $k$-th iteration of natural logarithm?

  • $\begingroup$ Is the tag "real-analysis" really appropriate for $k > 2$? :). $\endgroup$
    – Erick Wong
    Feb 3, 2014 at 16:33

1 Answer 1


$S(\alpha_1,\cdots,\alpha_k)$ converges if and only if there exists $j<k$ such that for all $1\leq i \leq j$, $\alpha_i = 1$ and $\alpha_{j+1}>1$.


We use the Cauchy Condensation Test, which states

A positive decreasing sequence $a_n$ converges if and only if $\sum 2^n a_{2^n}$ converges.

By this test, $S(\alpha)=\sum\limits_{n=3}^\infty \frac{1}{n (\log (n)^{\alpha})}$ converges iff $\sum\limits_{n=3}^\infty \frac{2^n}{2^n (n \log(2))^\alpha} = \frac{1}{(\log 2)^\alpha}\sum\limits_{n=3}^\infty \frac{1}{n^\alpha}$ converges if and only if $\alpha >1$.

Iterating this argument, we see that $S(1,\cdots,1,\alpha)$ converges if and only if $S(\alpha)$ converges. By the Comparison Test, $S(1,\cdots 1,\alpha_j,\cdots \alpha_k)$ converges if $S(1,\cdots,1,\alpha_j)$ converges.


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