Sequence is not in any $\ell^p$ space

Question

We know the sequence {$\frac{1}{ln(n)}$} such that $(n>=2)$ converges to $zero$ but is not in any $L_p$ space because of $$\sum_{n=2}^{\infty}\left|{\frac{1}{ln(n)}}\right|^p ={\infty}$$ for any $(p>1)$ . Now let we have the following sequences : $$x_n=\left(\frac{1}{\sqrt[n]{n}}-1\right)$$ $$y_n=\left(\sqrt[n]{n}-1\right)$$ Such that ($n>=1$) Clearly : $x_n$ and $y_n$ converge to $zero$ But .Do you think also they are not in any $L_p$ Space ( I mean that $$\sum_{n=1}^{\infty}\left|\frac{1}{\sqrt[n]{n}}-1\right|^p={\infty}??$$ $$\sum_{n=1}^{\infty}\left|\sqrt[n]{n}-1\right|^p={\infty}??$$ For any $p>1$). I think yes $$\sum_{n=1}^{\infty}|x_n|^p={\infty}$$ and $$\sum_{n=1}^{\infty}|y_n|^p={\infty}$$ For any $p>1$ Maybe I'm not correct , but I want the correct answer and the proof . With all my respect

Answer 1

First of all, the two sequences differ by a factor which is bounded away from both zero and $\infty$. Specifically we have $x_n = \frac{1-\sqrt[n]{n}}{\sqrt[n]{n}}=-\frac{1}{\sqrt[n]{n}} y_n$. So it's enough to study one or the other.

So we'll study $y_n$ here. We have

$$\sqrt[n]{n}-1 = n^{1/n}-1 = \exp \left ( \frac{\ln(n)}{n} \right )-1 > 1+\frac{\ln(n)}{n}-1=\frac{\ln(n)}{n}$$

where we obtained the inequality using the facts that $\exp$ is convex, $\exp(0)=1$, $\exp'(0)=1$. So $y_n$ is not summable as we can see by comparison with the harmonic series.