I am attempting to test the convergence for the following series for a final review:
$$\sum_{n=2}^{\infty} \frac{1}{n^{p} \ln(n)^{2}}, p>0$$
My classmates and I have tested the convergence for when $p=1$ and $p>1$; the series converges in both cases (Comparison Test). However, we are having problems for when $0<p<1$. Was wondering if we could get some hints for this third case.
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FlashKicks909
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2Your series in the title differs from that in the question. In the exponent on $\ln(n)$ equal to $p$ or $2$? – 2'5 9'2 Aug 07 '16 at 01:19
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I that (\log n )^2 ,that is, $(\log n)^2$ ? Or \log (n^2), that is, $\log (n^2)$ ? – DanielWainfleet Aug 07 '16 at 03:54
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Hint: Cauchy Condensation Test. – DanielWainfleet Aug 07 '16 at 03:54
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2Possible duplicate of On convergence of Bertrand series $\sum\limits_{n=2}^{\infty} \frac{1}{n^{\alpha}\ln^{\beta}(n)}$ where $\alpha, \beta \in \mathbb{R}$ or this. – Arnaud D. Dec 03 '19 at 11:09
2 Answers
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Edit: this answer addresses the original question, $\sum_{n=2}^{\infty}\frac{1}{n^p\ln(n)^p}$.
The series diverges when $p=1$ by the integral test.
When $0<p<1$, using the fact that $$ \lim_{x\to\infty}\frac{\ln(x)}{x^a}=0 $$ for any $a>0$, we have $$ n^p\ln(n)^p\leq n^pn^{ap}=n^{p(a+1)}$$ for all sufficiently large $n$ (depending on $a$). Choosing $a>0$ small enough that $p(a+1)<1$, it follows that $\sum_{n=1}^{\infty}\frac{1}{n^{p(a+1)}}$ diverges, and since $$ \frac{1}{n^p\ln(n)^p}\geq \frac{1}{n^{p(a+1)}} $$ for all large enough $n$, this shows that $\sum_{n=2}^{\infty}\frac{1}{n^p\ln(n)^p}$ diverges as well.
carmichael561
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Alternately, once you have the divergence for $p=1$ you can just use the comparison test against that series for $p\lt 1$. – Steven Stadnicki Aug 07 '16 at 00:53
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I apologize, but there was an error when I was transcribing the problem. – FlashKicks909 Aug 07 '16 at 01:18
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If the series is the series stated in the question (not the one in the title) then the series converges when $p=1$ (and the Integral Test can show that). – 2'5 9'2 Aug 07 '16 at 01:21
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If you are asking about $$\sum_{n=2}^{\infty}\frac{1}{n^p\ln(n)^2}$$ then take the integral $$\int_2^{\infty}\frac{1}{x^p\ln(x)^2}\,dx$$ and make the substitution $u=\ln(x)$ to get $$\int_{\ln(2)}^{\infty}\frac{1}{e^{(p-1)u}u^2}\,du$$ If $p<1$, then the integrand itself approaches $\infty$ as $u\to\infty$, so the integral diverges. So the series would diverge for $p<1$ by the Integral Test.
2'5 9'2
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