Find all values of real number p for which the series converges absolutely
$$\sum_{k=2}^{\infty} \frac{1}{k\, (\log{k})^p}$$
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Ron Gordon
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Amber
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3What have you tried? Do you know Cauchy's condensation test? Do you know the Integral test? – Pedro Apr 18 '13 at 21:26
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My book hasn't said anything about Cauchy's condensation test yet. I've tried just placing different general values in for p (p = 0, p >0, p < 0). I can see why p = 0 would not work, but I am having trouble seeing why it doesn't converge when p < 0 (according to the back of the book.) And I was thinking the if p > 0 it would converge because it would be similar to a p-series. I still don't quite understand the integral test. – Amber Apr 18 '13 at 21:36
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It is good your write that everytime you ask this kind of questions. – Pedro Apr 18 '13 at 21:39
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Possible duplicate of Convergence of the series $\sum \limits_{n=2}^{\infty} \frac{1}{n\log^s n}$ – Arnaud D. Dec 05 '19 at 10:37
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This is amenable to the well-known Cauchy condensation test that goes as follows:
If we have a non-negative decreasing sequence $\{a_n\}_{n=1,2,\ldots}^{\infty}$, then $$\displaystyle \sum_{n=2}^{\infty} a_n \text{ converges iff }\displaystyle \sum_{n=2}^{\infty} 2^na_{2^n} \text{ converges}$$
In your case, we have $a_n = \dfrac1{n \log^p(n)}$.
Hence,$$\sum_{n=2}^{\infty} 2^na_{2^n} = \sum_{n=2}^{\infty} \dfrac{2^n}{2^n (n \log2)^p} = \dfrac1{(\log2)^p}\sum_{n=2}^{\infty} \dfrac1{n^p}$$ which converges for $p>1$ and diverges for $p \leq 1$.
