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I am having trouble showing that $\ell_1\subset\ell_2\subset c_0\subset\ell_\infty$, where $c_0$ is the subset of $\ell_\infty$ consisting of all sequences that converge to 0. Specifically, I am having trouble showing that $\ell_2\subset c_0$.

Here is my my entire proof: Trivially, $c_0\subset \ell_\infty$ by definition of $c_0$. Since $\|x\|_1\geq\|x\|_2$, the set of sequences $x_j\in\mathbb{R}$ such that $\sum_j|x_j|<\infty$ is smaller than the set of sequences such that $\sum_j|x_n|^2$. The subset is proper because $x_n=\frac{1}{n}$ converges in $\ell_2$ (to $\frac{\pi^2}{6}$) but not in $\ell_1$. Thus $\ell_1\subset\ell_2$ and $c_0\subset\ell_\infty$. We just need to show that $\ell_2\subset c_0$.

But since $x_n=\sum|\frac{1}{n}|^2$ doesn't converge to 0, this means that $\ell_2$ cannot be a subset of $c_0$. Where have I gone wrong? Do I lack a fundamental understanding of $\ell_p$ spaces?

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    $\begingroup$ Try to show that if $\sum a_n < \infty$ then $\lim_{n\to \infty} a_n = 0$. Hint: What happens to the sum $\sum a_n$ if either the limit $\lim_{n \to \infty} a_n \neq 0$ or does not exist? $\endgroup$ – Vitor Borges Sep 20 '16 at 23:47
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You're confusing the sum of the series $\sum_{n=1}^{\infty}a_n$ with the limit of the sequence $\lim_{n\to\infty}a_n$. While the sum may be non-zero, it's a standard fact that if $\sum_{n=1}^{\infty}a_n$ converges then $\lim_{n\to\infty}a_n= 0$.

In particular, if $\sum_{n=1}^{\infty}|x_n|^2<\infty$, then taking $a_n=|x_n|^2$ shows that $\lim_{n\to\infty}|x_n|^2=0$, hence also $\lim_{n\to\infty}x_n=0$.

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  • $\begingroup$ So, for example , $x=1+\frac{1}{n}$ is in $\ell_\infty$ because it converges, but not in $c_0$, because it doesnt converge to 0? $\endgroup$ – Matt G Sep 20 '16 at 23:48
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    $\begingroup$ Sure. But a sequence doesn't need to be converge to be in $\ell^{\infty}$, it just needs to be bounded. $\endgroup$ – carmichael561 Sep 20 '16 at 23:55
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    $\begingroup$ So $(-1)^n$ would be in $\ell_\infty$? $\endgroup$ – Matt G Sep 20 '16 at 23:56
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    $\begingroup$ Yes, that's correct. $\endgroup$ – carmichael561 Sep 20 '16 at 23:56

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