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I need to find a sequence which converges to $0$ but is not in any space $\ell^p$, where $1 \leq p < +\infty$.

And, I need to find a sequence which is in every space $\ell^p$ with $p > 1$ but which is not in $\ell^1$.

As a recap, for any $p$ such that $1 \leq p < +\infty$, we define the space $\ell^p$ to be the set of all sequences $x \colon= (\xi_i)$ of complex numbers for which the series $\sum |\xi_i|^p $ converges to a finite (real) value, with the metric $d$ defined as follows: $$ d(x,y) \colon= \sum_{i=1}^\infty |\xi_i - \eta_i|^p $$ for all $x \colon= (\xi_i)$, $y \colon= (\eta_i)$ in $\ell^p$.

My work:

The sequence $\xi_i \colon= i^{-1/(p+1)} $ obviously converges to $0$, but we have $|\xi_i|^p = i^{-p/(p+1)}$, and $0 < p/(p+1) < 1$; so the series $\sum i^{-p/(p+1)}$ diverges to $+\infty$. But can we find a single sequence that works for every $p$, where $1 \leq p < +\infty$?

And similarly, for the sequence $\xi_i \colon= i^{-2(p-1)/p}$, we have $|\xi_i|^p = i^{-2(p-1)}$, and $2(p-1) = 0$ for $p=1$ and $2(p-1) > 1$ for $p> 1$; so the series $\sum i^{-2(p-1)}$ diverges to $+\infty$ for $p=1$ and this series converges (to some finite, non-negative real value) for $p > 1$. But, as before, can we find a single sequence that works for every $p$, where $1 \leq p < +\infty$?

Saaqib Mahmood
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  • When you say that sequence $\xi_i$ converges to $0$, do you mean $\forall \varepsilon > 0 \exists N \forall n. n> N \Rightarrow |x_i|<\varepsilon$? – Antoine Sep 17 '14 at 15:37
  • Yes of course, I mean convergence with respect to the usual metric on $\mathbb{R}$. – Saaqib Mahmood Sep 18 '14 at 01:25

1 Answers1

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Idea: Let $\xi_n\in\mathbb{R}$. If $$\lim_{n\to \infty} \frac{\xi_n}{n^{-\alpha}} < \infty$$ for any $\alpha > 0$, then there exists $N$ such that $\xi_n<Cn^{-\alpha}$ for all $n>N$, hence $$\sum_n |\xi_n|^\frac{2}{\alpha}<\infty\text{.}$$

Therefore, $|\xi_n|$ must be very slowly decreasing sequence. Very slowly increasing is $\log n$, therefore we can try with $$\xi_n = \frac{1}{\log (1+n)}\text{,}$$

since $\log 1 = 0$.

Let be $p\geq 1$. Then, 'obviously', there is a $N_p$ such that $\log (n+1)<(n+1)^\frac{1}{1+p}$ for all $n\geq N_p$, hence

$$\sum_{n=1}^\infty |\xi_n|^p\geq\sum_{n=N_p}^\infty |\xi_n|^p \geq \sum_{n=N_p}^\infty (n+1)^{-\frac{p}{p+1}} = \infty\text{.}$$

Antoine
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  • Antoine, thank you for your answer, but I'm afraid your argument isn't clear to me. You might want to edit your answer for some possible typing mistakes. – Saaqib Mahmood Sep 18 '14 at 02:31
  • @SaaqibMahmuud Which argument(s) is(are) not clear? Also, sorry, but I cannot find any typing mistakes. – Antoine Sep 18 '14 at 19:57