1
$\begingroup$

I am having trouble determining convergence/ divergence of the following series:

$\sum\limits_{n=2}^{\infty} \frac{1}{(\log(n))^{1/n}}$

When I apply the root and ratio tests, I find in both cases the limit 1 (which means that the test is inconclusive).

Please help

  • 1
    Hint: What is $\lim\limits_{n\rightarrow\infty} \bigl(\log (n)\bigr)^{1/n}$?2012-12-12

2 Answers 2

2

Just use the limit test. It is $\log(n)^{1/n}\longrightarrow 1\neq 0$, so the series cannot converge.

  • 1
    @DavidMitra note that's actually the contrapositive of the usual theorem that's taught ($\sum a_n$ converges $\implies a_n \to 0$).2014-06-02
0

We can use the fact that if $a_n>0$ for all $n\ge1$ and the sequence $\frac{a_{n+1}}{a_n}$ converges in $[0,\infty]$, then $ \lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n} $ (see this question).

We have that $ \lim_{n\to\infty}\biggl(\frac1{\log n}\biggr)^{1/n}=\lim_{n\to\infty}\frac{\frac1{\log(n+1)}}{\frac1{\log n}}=\lim_{n\to\infty}\frac{\log n}{\log n+\log(1+1/n)}=1. $ Hence, the series diverges.