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

Question

I need to test for convergence and absolute convergence this sum: $\sum_{n=2}^{\infty} \frac {1}{\ln^n(\frac{1}{n})} $.

Thank you for your help.

Is $\ln^n$ the $n$th power or the $n$th iterated logarithm? – Bart Michels May 27 '18 at 15:33 — Bart Michels, May 27 '18 at 15:33

score 2 · Accepted Answer · answered May 27 '18 at 15:34

2

Hint: $\displaystyle\frac1{\ln^n\left(\frac1n\right)}=\frac1{\bigl(-\ln(n)\bigr)^n}=\frac{(-1)^n}{\ln^n(n)}.$

answered May 27 '18 at 15:34

And how does this help to determine the absolute convergence? – Mark Viola May 27 '18 at 16:11
1

It doesn't, but it's useful to prove that it converges. – José Carlos Santos May 27 '18 at 16:14

score 2 · Answer 2 · answered May 27 '18 at 15:37

2

HINT:

Note that for $n>e^2$,

$$\left|\frac{1}{\log^n(1/n)}\right|\le \frac1{2^n}$$

answered May 27 '18 at 15:37

Mark Viola

score 1 · Answer 3 · answered May 27 '18 at 15:57

We have that

$$\frac1{\ln^n\left(\frac1n\right)}=\frac1{\bigl(-\ln(n)\bigr)^n}=\frac{(-1)^n}{\ln^n(n)}$$

and by Cauchy condensation test we have

$$\sum \left|\frac{(-1)^n}{\ln^n(n)}\right|\le \sum \frac{2^n}{(\ln (2^n))^{2^n}}=\sum \frac{2^n}{n^{2^n}(\ln 2)^{2^n}}$$

3 Answers3