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Analyze the convergence of $$\sum\frac{1}{n(\log(n))^c}.$$

I know that it diverges for $c\leq1$ by using comparision test and integral test for convergence through the $$\int_{2}^{\infty}\frac{1}{x\log x}dx= \log(\log(\infty))-\log(\log2)$$

I also have a gut feeling that it converges for $c>1$ but not able to prove so. Please try explaining through less exotic methods by not resorting to weird convergence test theorem. Thanks!

José Carlos Santos
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markovchain
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  • Integral test can be used for any $c$. Just make the change of variable $y=\ln x$ – Kavi Rama Murthy Jun 18 '18 at 08:12
  • the solution using cauchy's condensation test is interesting and that's not present in the duplicate problem, therefore we may retain this question as well – markovchain Jun 18 '18 at 11:34
  • Another one: https://math.stackexchange.com/q/9911 (which covers both solutions, plus generalizations). Also https://math.stackexchange.com/q/1628839, https://math.stackexchange.com/q/641310. – Martin R Jun 18 '18 at 11:48

2 Answers2

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By Cauchy condensation test

$$ 0 \ \leq\ \sum_{n=1}^{\infty} f(n)\ \leq\ \sum_{n=0}^{\infty} 2^{n}f(2^{n})\ \leq\ 2\sum_{n=1}^{\infty} f(n)$$

we have that

$$\sum\frac{1}{n(\log(n))^c}\le \sum\frac{2^n}{2^n(\log(2^n))^c}=\sum\frac1{n^c\log^c 2}$$

user
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2

The integral test also works: by substitution, we have $$\int_{2}^\infty \frac{\mathrm d t}{t\,\ln^ct}=-\frac 1{(c-1)\ln^{c-1}t}\Biggr|_2^\infty=0+\frac1{(c-1)\ln^{c-1}2}\quad\text{ if }\;c>1.$$

Bernard
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