Let $p > 1$. Is the series $$\sum_k\left| \frac{1}{\ln{(k+1)}}\right|^p$$ divergent?
This is in the context of finding a sequence that goes to zero but is not an element of $l^p$ space. I saw in another post that this can be proved to be divergent by the limit comparison test using $x_k = \frac{1}{k}$ as the testing sequence. I'm attempting to show this below, assuming that $1 + k > e$.
Let $y_k = \left| \frac{1}{\ln{(k+1)}}\right|^p$, and $x_k = \frac{1}{k}$, then
\begin{align} \lim_{k\to \infty}\left|\frac{y_k}{x_k}\right| &= \lim \frac{k}{\left[\ln(k+1)\right]^p} \\ \\ &\overset{l'h}{=} \frac{1}{p}\lim \frac{k+1}{\left[\ln(k+1)\right]^{p-1}} \\ \\ &\overset{l'h}{=} \frac{1}{p(p-1)}\lim \frac{k+1}{\left[\ln(k+1)\right]^{p-2}} \\ \\ &\, \, \, \vdots \\ \\ &\overset{l'h}{=} \frac{1}{p(p-1)\cdots(p-n)}\lim \frac{k+1}{\left[\ln(k+1)\right]^{p-n}} \end{align}
Since $p > 1$ is fixed, there must exist an $n \in \mathbb{N}$ such that $p - n \leq 0$. For that value of $n$ let $p - n = -a$. Then
\begin{align} \frac{1}{p\cdots(p-n)}\lim \frac{k+1}{\left[\ln(k+1)\right]^{p-n}} &= \frac{1}{p\cdots(p-n)}\lim \frac{k+1}{\left[\ln(k+1)\right]^{-a}} \\ \\ &=\frac{1}{p\cdots(p-n)}\lim (k+1){\left[\ln(k+1)\right]^{a}} \\ \\ &\longrightarrow \infty \quad \text{ as } \quad k \longrightarrow \infty \end{align}
Therefore, by my recollection of the limit comparison test, this proves that the sequence term in the numerator $y_k$ must be larger than the denominator term $x_k$ at some value of $k$, and then forever more. Basically, the top is bigger than the bottom and the bottom has a divergent series, so the top must represent the terms of a divergent series.
Is this proof sufficient? And/or does anybody have simpler proof lying around? Also, I tried the integral test on this but is seemed difficult, so is it possible that way?
Here's a simple proof using the Cauchy Condensation Test.
Observe \begin{align} \sum_{k=2}\frac{1}{|\ln k|^p} \end{align} is convergent if and only if \begin{align} \sum_{\ell=1}2^\ell \frac{1}{|\ln 2^\ell|^p} \end{align} is convergent. However, for the latter sum, we see that \begin{align} \sum_{\ell=1} \frac{2^\ell}{\ell^p |\ln 2|^p} = \frac{1}{|\ln 2|^p}\sum_{\ell=1} \frac{2^\ell}{\ell^p} \end{align} which is clearly divergent for any $p>1$ since $2^\ell/\ell^p \rightarrow \infty$ as $\ell\rightarrow \infty$.