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 $p>1$.

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.