Seeking reasoning and verification that $$\sum_{n=1}^\infty \left(\frac{1}{(n+1)\ln^2 (n+1)}\right)^r$$ diverges for $0<r<1$.
Referring to a thread from 2016:
I verified that the sum converges for $r=1$ using the integral test, but am having a difficult time verifying its divergence for $0<r<1$.
My apologies if this is obvious. My available resources and textbooks lack clarity on the subject.
Claim 1: For any $\beta\in(0,\infty)$, there exists $n_{0}$ such that $\ln n\leq n^{\beta}$ whenever $n\geq n_{0}$.
Proof: Let $\beta\in(0,\infty)$ be given. By L Hospital rule, \begin{eqnarray*} & & \lim_{x\rightarrow\infty}\frac{\ln x}{x^{\beta}}\\ & = & \lim_{x\rightarrow\infty}\frac{\frac{1}{x}}{\beta x^{\beta-1}}\\ & = & \frac{1}{\beta}\lim_{x\rightarrow\infty}\frac{1}{x^{\beta}}\\ & = & 0. \end{eqnarray*} Therefore, there exists $x_{0}$ such that $\frac{\ln x}{x^{\beta}}<\frac{1}{2}$ whenever $x\geq x_{0}$. In particular, $\ln n\leq n^{\beta}$ whenever $n\in\mathbb{N}\cap(x_{0},\infty)$.
For your problem: Let $r\in(0,1)$ be given. Choose $\beta\in(0,\frac{1-r}{2r})$. Choose $n_{0}\in\mathbb{N}$ such that $\ln n\leq n^{\beta}$ whenever $n\geq n_{0}$. Denote $\alpha=(2\beta+1)r$. Note that $0<\alpha<1$.
For $n\geq n_{0}$, we have that \begin{eqnarray*} & & n^{r}\left(\ln n\right)^{2r}\\ & \leq & n^{r}\cdot\left(n^{\beta}\right)^{2r}\\ & = & n^{(2\beta+1)r}\\ & = & n^{\alpha}. \end{eqnarray*} It follows that \begin{eqnarray*} & & \sum_{n=2}^{\infty}\left(\frac{1}{n\ln^{2}n}\right)^{r}\\ & \geq & \sum_{n=n_{0}}^{\infty}\frac{1}{n^{\alpha}}\\ & = & \infty. \end{eqnarray*} Hence, $\sum_{n=2}^{\infty}\left(\frac{1}{n\ln^{2}n}\right)^{r}$ diverges.
All that is needed is that, for $0 < c < 1$, $\lim_{x \to \infty} \dfrac{\ln(x)}{x^c} =0 $.
This has been proved here many times, often by me, so I will leave this to you.
Since $0 < r < 1$, let $c = \dfrac{1-r}{2} $ so $0 < c < 1$, $r=1-2c$, and $c < 1-r$.
Then, writing $n$ for $n+1$ since it doesn't matter, we want to show that $(n \ln^2(n))^r \lt n^s$ for some $0 < s < 1$, since $\sum_{n=1}^{\infty} \dfrac1{n^s}$ diverges.
$\begin{array}\\ (n \ln^2(n))^r &=(n \ln^2(n))^{1-2c}\\ &=n^{1-2c} \ln^{2(1-2c)}(n)\\ &\lt n^{1-2c} \ln^{2}(n)\\ &\lt n^{1-c} n^{-c}\ln^{2}(n)\\ &\lt n^{1-c} \dfrac{\ln^{2}(n)}{n^c}\\ &\lt n^{1-c} \qquad\text{for large enough }n\\ \text{so}\\ \dfrac1{(n \ln^2(n))^r} &\gt \dfrac1{n^{1-c}} \qquad\text{for large enough }n\\ \end{array} $
and this sum of this diverges.
No te that this works for $\dfrac1{(n \ln^b(n))^r}$ for any $b > 0$.
We can show given series diverges using Bertrand's test of convergence. From the series, $a_n=\frac{1}{(n+1)ln^2(n+1)}$ Now let's look at the following limit, where $0<r<1$: $$\lim_{n\to\infty}\left(\left(n\left(\frac{\left(\frac{1}{(n+1)\ln^2(n+1)}\right)^r}{\left(\frac{1}{(n+2)\ln^2(n+2)}\right)^r}-1\right)-1\right)\ln{n}\right)=\lim_{n\to\infty}\left(\left(n\left(\left(\frac{(n+2)ln^2(n+2)}{(n+1)ln^2(n+1)}\right)^r-1\right)-1\right)\ln{n}\right)=\lim_{n\to\infty}{((n(1-1)-1)\ln n)}=\lim_{n\to\infty}{(-\ln{n})}=-\infty<1$$
Thus, using the Bertrand's test of convergence we can conclude that the series $\displaystyle\sum_{n=1}^\infty \left(\frac{1}{(n+1)\ln^2 (n+1)}\right)^r$ diverges for $0< r<1$. You can easily show that it also diverges for all $r\le0$, meaning the series diverges for all $r<1$.
As indicated in the comment section we can conveniently use Cauchy's condensation test. We write the series as \begin{align*} \sum_{n=1}^{\infty}\left(\frac{1}{(n+1)\ln^2(n+1)}\right)^r=\sum_{n=2}^{\infty}\left(\frac{1}{n\ln^2(n)}\right)^r \tag{1} \end{align*}
According to Cauchy's condenation test the right-hand side of (1) converges iff the following series converges. Given $0<r<1$, we obtain \begin{align*} \color{blue}{\sum_{n=2}^{\infty}}\color{blue}{2^n\left(\frac{1}{2^n\ln^2\left(2^n\right)}\right)^r} &=\sum_{n=2}^{\infty}2^{n(1-r)}\frac{1}{\left(\ln(2^n)\right)^{2r}}\\ &=\sum_{n=2}^{\infty}2^{n(1-r)}\frac{1}{\left(n\ln (2)\right)^{2r}}\\ &\,\,\color{blue}{=\frac{1}{\left(\ln (2)\right)^{2r}}\sum_{n=2}^{\infty}2^{n(1-r)}\frac{1}{n^{2r}}\to\infty}\tag{2} \end{align*} Since (2) diverges, the divergence of (1) follows.
