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Consider the series $$\sum_{n \geq 2} \frac{1}{n^p \ln^qn}$$ Prove that:

  • The series converges if $p > 1$ (and any $q$), or if $p = 1$ and $q > 1$.
  • The series diverges if $p < 1$ (and any $q$), or if $p = 1$ and $q \leq 1$.

Immediately I know that we must have $p > 0$, otherwise the general term won't go to zero.

The ratio test fails, since if we call $x_n = \frac{1}{n^p \ln^qn}$, we get that: $$\frac{x_{n+1}}{x_n} = \frac{n^p \ln^q n}{(n+1)^p \ln^q(n+1)} = \left(\frac{n}{n+1}\right)^p \left(\frac{\ln n}{\ln(n+1)}\right)^q \to 1 \cdot 1 = 1$$

Since $n^{p/n} = e^{(p \ln n)/n}$ and $\ln^{q/n}n = e^{(q \ln \ln n)/n}$, the root test also fails: $$\sqrt[n]{|x_n|} = \sqrt[n]{\frac{1}{n^p \ln^qn}} = \frac{1}{n^{p/n} \ln^{q/n}n} \to \frac{1}{1 \cdot 1} = 1$$ so, no good. I hardly think that the integral test will help here. I thought the idea was to get some condition on $p$ and $q$ by using the above tests.

Then, I'm left with comparing it with $1/n^2$ or something like it, but I'm a little lost about how to go about it. Can someone give me a hand? Thanks.

Nosrati
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Ivo Terek
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2 Answers2

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The Cauchy's condensation test might not be widely recognized among students. In this context, we employ the comparison test and the integral test to assess the convergence of this series. To start, we utilize L'Hôpital's rule, yielding the following result: $$ \lim_{x\rightarrow+\infty}\frac{\ln^\alpha(x)}{x^\beta}=0 \quad (1) $$ This result holds true for all real values of $\alpha$ and positive real values of $\beta$. We proceed by examining three distinct cases:

Case 1. When $p>1$

Consider $r\in (1,p)$. According to (1), we can establish $$ \frac{1}{n^p\ln^q(n)}:\frac{1}{n^r}=\frac{\ln^{-q}(n)}{n^{p-r}}\longrightarrow 0. $$ As the series $\sum_{n=2}^{\infty}\frac{1}{n^r}$ converges, we can conclude that the series $\sum_{n=2}^{\infty}\frac{1}{n^p\ln^q(n)}$ is also convergent.

Case 2. $p<1$

Consider $r\in (p,1)$. Following (1), we can establish $$ \frac{1}{n^p\ln^q(n)}:\frac{1}{n^r}=\frac{\ln^{-q}(n)}{n^{p-r}}\longrightarrow +\infty. $$ As the series $\sum_{n=2}^{\infty}\frac{1}{n^r}$ diverges, we can conclude that the series $\sum_{n=2}^{\infty}\frac{1}{n^p\ln^q(n)}$ is also divergent.

Case 3. $p=1$

We can follow the final part of Ivo Terek's answer to reach the desired conclusion.

Blind
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2

With the suggestions from the comments, looking up the links given, etc, I'll post my solution. Since $\frac{1}{n^p \ln^qn}$ is decreasing and non-negative, we use Cauchy's condensation test. We have to look at: $$\sum_{n \geq 2} \frac{2^n}{(2^n)^p (\ln 2^n)^q} = \sum_{n \geq 2} \frac{1}{\ln^q2} \frac{2^n}{2^{np}n^q} = \frac{1}{\ln^q2} \sum_{n \geq 2} \frac{2^n}{2^{np}n^q} = \frac{1}{\ln^q 2} \sum_{n \geq 2} \frac{2^{n(1-p)}}{n^q}$$

Now we use the ratio test for $\sum_{n \geq 2} \frac{2^{n(1-p)}}{n^q}$. We have that:

$$\frac{2^{(n+1)(1-p)}}{(n+1)^q}\frac{n^q}{2^{n(1-p)}} = 2^{(n+1)(1-p) - n(1-p)} \left(\frac{n}{n+1}\right) \to 2^{1-p}$$

So, if $p = 1$, the test fails. If $p > 1$, $2^{1-p} < 1$, and the series converges. In the same way, $p < 1$ gives us $2^{1-p} > 1$ and the series diverges.

Now, we only have to look at the case $p = 1$. For this, we can go back to the original series $\sum_{n \geq 2} \frac{1}{n \ln^qn}$. Now, we use the integral test. Finally, we have: $$\int_2^{+\infty} \frac{1}{x \ln^q x} \ \mathrm{d}x = \left.\frac{(\ln x)^{-q+1}}{-q+1}\right|_2^{+\infty} = \frac{1}{-q+1}\lim_{\alpha \to +\infty} \left((\ln \alpha)^{-q+1} - (\ln 2)^{-q+1}\right)$$

This is valid if $q \neq 1$. Since $(\ln 2)^{-q+1}$, we only have to look at $(\ln \alpha)^{-q+1}$. It will blow up if $-q+1 < 0$, that is, $q < 1$, so the series will diverge. On the other hand, if $q > 1$, $(\ln \alpha)^{-q+1} \to 0$ and the series will converge.

And if $p = q = 1$, we have: $$\int_2^{+\infty} \frac{1}{x \ln x} \ \mathrm{d}x = \lim_{\alpha \to +\infty} \ln(\ln \alpha) - \ln(\ln 2) = +\infty.$$

Ivo Terek
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