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Let $\ell^p:=\{(x_n)|\|x_n\|_{\ell^p}<\infty\}$ be the space of sequences with finite $\ell^p$-norm.

Show that the sequence $x_1:=1$, $x_k:=\frac{1}{\log k}$, $k\geq 2$ converges to $x=0$, but $x\notin\ell^p$ for $p\geq 1$.

To show that the sequence converges in $\ell^p$ we have to show that $\|x_n-0\|_{\ell^p}\to 0$. So basically we have to consider the series $\|x_n\|^p_{\ell^p}=\sum_{k=1}^\infty |x_k|^p$. Since a finite number of summands does not change the limit, we can consider the series

$$\sum_{k=2}^\infty \left|\frac{1}{\log k}\right|^p = \sum_{k=1}^\infty \left|\frac{1}{\log(k+1)}\right|^p$$

Now I have to show that the series converges to $0$. For $p>1$ we can use the estimate $\log(k)\leq k-1$ for $k>0$ and hence $$\sum_{k=1}^\infty \left|\frac{1}{\log(k+1)}\right|^p\leq\sum_{k=1}^\infty \left|\frac{1}{k}\right|^p\to 0\quad\mathrm{since\ }p>1$$

But how do I show the convergence for $p=1$? Does the series even converge in this case? And how do I show that $0\notin\ell^p$ (or is this simply because $0$ is a real number and not a sequence?).

Any advice is appreciated, thanks!

Edit: I just realized I made a mistake using the estimate $\log(k)\leq k-1$. From this estimate we get

$$\sum_{k=2}^\infty \left|\frac{1}{\log(k)}\right|^p>\sum_{k=2}^\infty \left|\frac{1}{k-1}\right|^p \quad\begin{cases} \to\infty, & p=1\\ <\infty, & p>1\end{cases}$$

So, having the comments below in mind, for $p=1$ we get that the series does not lie in $\ell^1$. However I still don't know how to show $x_k\notin\ell^p$ for $p>1$.

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  • $\begingroup$ Your notation is unclear: convergence in $l^p$ means convergence of a sequence of sequences. But you write only about one fixed sequence of scalars... $\endgroup$ – daw Apr 9 '14 at 7:24
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    $\begingroup$ The question did not ask you to show $0\notin \ell^p$. I think this question just show you that there are sequence $(x_n)$ which converges to 0 (in elementary analysis sense), but the sequence itself is not in $\ell^p$ for all $p\geq 1$. $\endgroup$ – user99914 Apr 9 '14 at 7:25
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    $\begingroup$ okay, I think I might've gotten the question all wrong then... So basically I have to show that $x_k\to 0$, which is obvious since $\log(k)\to\infty$ for $n\to\infty$, but $(x_k)\notin\ell^p$? $\endgroup$ – dinosaur Apr 9 '14 at 7:33
  • $\begingroup$ I made an edit to my post, having realized a mistake I made using the estimate $\log(k)\leq k-1$. However I'm stuck proving that $(x_k)\notin\ell^p$ for $p>1$. $\endgroup$ – dinosaur Apr 9 '14 at 7:39
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    $\begingroup$ Hint: You can use the fact: for all $p\geq 1$, there is $N_p$ large such that $(\log n)^p \leq n$ (Can be checked by L'Hospital rule) $\endgroup$ – user99914 Apr 9 '14 at 7:45
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To show that $(x_n)\not\in \ell^p$, you can use Cauchy-condensation. Then you have: $$\sum_{n=1}^{\infty}\left(\frac{2^n}{\log(2^n)}\right)^p=\sum_{n=1}^{\infty}\left(\frac{2^n}{n\log(2)}\right)^p=\frac{1}{\log(2)^p}\sum_{n=1}^{\infty}\left(\frac{2^{n}}{n}\right)^p=\infty$$

since $\lim_{n\to \infty}\frac{2^n}{n}\neq0.$

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  • $\begingroup$ I have never heard of Cauchy condensation so far, but this is really great! I will catch up on that! Thanks! $\endgroup$ – dinosaur Apr 9 '14 at 8:09
  • $\begingroup$ @dinosaur I believe there is a wikipedia entry about it. $\endgroup$ – Vincent Boelens Apr 9 '14 at 8:11

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