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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

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    $\begingroup$ Hint: What is $\lim\limits_{n\rightarrow\infty} \bigl(\log (n)\bigr)^{1/n}$? $\endgroup$ – David Mitra Dec 12 '12 at 13:18
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Just use the limit test. It is $\log(n)^{1/n}\longrightarrow 1\neq 0$, so the series cannot converge.

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  • $\begingroup$ The limit test states that if $\lim_{n \to \infty}\frac{a_n}{b_n}=1$ and $\sum b_n$ converges then $\sum a_n$ converges. How did you use it here? $\endgroup$ – user43758 Dec 12 '12 at 13:41
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    $\begingroup$ @user43758 He is not referring to that test, but rather the following: If $\lim_{n\rightarrow\infty} a_n\ne 0$, then the series $\sum_{n=1}^\infty a_n$ diverges. $\endgroup$ – David Mitra Dec 12 '12 at 13:43
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    $\begingroup$ @DavidMitra note that's actually the contrapositive of the usual theorem that's taught ($\sum a_n$ converges $\implies a_n \to 0$). $\endgroup$ – DanZimm Jun 2 '14 at 9:44
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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.

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