Divergence of $\sum_{n\geq 2} \frac{1}{\ln^p n}$ for $1Can anyone help me to prove that $(x_n)\notin l_p$ with $x_n=\frac{1}{\ln^p n}$? Suppose $1< p<\infty$.
 A: By the integral test for convergence, you may write, for all $N\geq3$,
$$
\int_3^N \frac{1}{\ln^p x} dx \leq \sum_{3\leq n \leq N} \frac{1}{\ln^p n}
$$ but, as $N \to \infty$,
$$
\int_3^N \frac{1}{\ln^p x} dx=\int_{\ln 3}^{\ln N} \frac{e^t}{t^p } dt\geq\int_1^{\ln N} e^t dt=N \, \to +\infty
$$
showing the divergence of the series $\displaystyle \sum_{n\geq 2} \frac{1}{\ln^p n}$, $1<p<\infty$.
A: Using the Cauchy condensation test, we have that
\begin{align*}
\sum_{n=1}^\infty 2^n x_{2^n} = \sum_{n=1}^\infty 2^n \dfrac{1}{\left(\log(2^n)  \right)^p} &=\sum_{n=1}^\infty \dfrac{2^n}{n^p}\dfrac{1}{\left( \log(2) \right)^p}
\end{align*}
which clearly diverges, hence the original series diverges
A: You can also use that, if we fix $\alpha>0
 $ then $n^{\alpha}
 $ is bigger than $\log^{p}\left(n\right),\,\forall p>0
 $ and a sufficiently large $n$, then $$n^{\alpha}>\log^{p}\left(n\right)
 $$ and so if we fix $\alpha<1
 $ we have, for a sufficiently large $n
 $, $n\geq N
 $ say, $$\sum_{n\geq N}\frac{1}{n^{\alpha}}<\sum_{n\geq N}\frac{1}{\log^{p}\left(n\right)}
 $$ and obviously the LHS diverges.
