Proving convergence and divergence of series

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

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!

 

[WWGD]

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.

 

[mathguy_1995]

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.

 

[pasmith]

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.