Proving convergence and divergence of series

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Homework Statement
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Relevant Equations
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For this problem,
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Let ##a_n = \frac{1}{n(\ln n)^p}##

##b_n = \frac{1}{(n \ln n)^p} = \frac{1}{(n^*)^p}##

We know that ##\sum_{2 \ln 2}^{\infty} \frac{1}{(n^*)^p}## is a p-series with ##n^* = n\ln n##, ##n^* \in \mathbf{R}##

Assume p-series stilll has the same property when ##n^* \in \mathbb{R}## instead of ##n \in \mathbb{N}##

This implies tht P-series is convergent when ##p > 1## and divergent when ##p \leq 1##

However, if we consider ##n^* = n \ln n## ##n^* \in \mathbb{N}## then ##\frac{1}{n (\ln n)^p} \leq \frac{1}{(n^*)^p}## when ##n \geq N##

Thus since ##\sum_{n = 2}^{\infty} \frac{1}{(n^*)^p}## converges for ##p > 1## then ##\sum_{n = 2}^{\infty} \frac{1}{n(\ln n)^p}## converges for ##p > 1## by comparison test

Since ##\sum_{n = 2}^{\infty} \frac{1}{(n^*)^p}## diverges for ##p \leq 1## then ##\sum_{n = 2}^{\infty} \frac{1}{n(\ln n)^p}## this implies that the other series diverges for ##p \leq 0## by comparison test

However, does anybody please know whether my proof is correct? I’m also not sure why author did not consider ##0 < p \leq 1##.

Thanks!
 
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For the first part, notice too, if ##p<0##, you can move the denominator to the top and consider more " standard" methods like the ratio test. But valid question on ##p \in (0,1)##. Let me think it through.
 
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Hello! Another quite useful tool to prove this is the Cauchy Condensation Test (##\sum_{k=1}^{\infty}a_k## is convergent if and only if ##\sum_{k=0}^{\infty}2^ka_{2^k}## is convergent). In this example, ##a_k=\frac{1}{k(\ln k)^p}##. Then, $$2^ka_{2^k}=2^k\frac{1}{2^k\left(\ln 2^k\right)^p}=\frac{1}{\left(\ln 2\right)^p}\frac{1}{k^p}$$ and
$$\sum_{k=2}^{\infty}2^ka_{2^k}=\frac{1}{\left(\ln 2\right)^p}\sum_{k=2}^{\infty}\frac{1}{k^p}.$$
Now, the problem is reduced to the convergence of ##\sum_{k=2}^{\infty}\frac{1}{k^p}##. For what values of ##p## does this last series converge? The Cauchy condensation test can give an answer to this question as well. Now, if ##b_k=\frac{1}{k^p}##,
$$2^kb_{2^k}=2^k\frac{1}{\left(2^k\right)^p}=\frac{1}{\left(2^k\right)^{p-1}}=\frac{1}{\left(2^{p-1}\right)^k}.$$
Now, the series ##\sum_{k=2}^{\infty}\frac{1}{\left(2^{p-1}\right)^k}## is a geometric series, hence it converges if and only if ##\frac{1}{2^{p-1}}<1## which means that it converges if ##p> 1## and diverges if ##\frac{1}{2^{p-1}}>1##, hence, when ##p\leq 1##.
I hope that this method gives a new perspective to your problem and ultimately helps.
 
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The intended method is the integral comparison test: if the a_n are positive and strictly decreasing for n \geq n_0, and f: [n_0, \infty) \to [0, \infty) is strictly decreasing such that f(n) = a_n for all n \geq n_0, then \sum_{n=n_0}^\infty a_n converges if and only if \int_{n_0}^\infty f(x)\,dx converges.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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