Divergence of $\sum_{n\geq 2} \frac{1}{\ln^p n}$ for $1<p\leq \infty$

Question

Can anyone help me to prove that $(x_n)\notin l_p$ with $x_n=\frac{1}{\ln^p n}$? Suppose $1< p<\infty$.

Answer 1

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 $\displaystyle \sum_{n\geq 2} \frac{1}{\ln^p n}$, $1<p<\infty$.

Answer 2

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

Answer 3

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.