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I want to show $c_{00}$ is dense in $\ell_2$. But I stuck in some calculus part.

And my attempt: Equip $c_{00}$ with $\|\cdot\|_2$. Let $x=(x_n)\in\ell_2$. The sequence $(y_n)=\{x_1,x_2,\cdots,x_n,0\cdots\}$ is a sequence in $c_{00}$. To show $c_{00}$ is dense in $\ell_2$, we only need to show that $y_n\rightarrow x$ in $\|\cdot\|_2$. And we have $\|y_n-x\|_2=(\sum_{k=n+1}^{\infty}|x_k|^2)^{\frac{1}{2}}.$ Let $\epsilon>0$. And I want to show $\|y_n-x\|_2<\epsilon$. It suffices to show $\sum_{k=N}^{\infty}|x_k|^2<\epsilon$ for some $N\in\mathbb{N}$. That is where I am stuck. I know we have $\sum_{k=N}^{\infty}|x_k|<\infty$. But how to get $\sum_{k=N}^{\infty}|x_k|^2<\epsilon$ for some $N\in\mathbb{N}$?

Thank you in advance!

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  • $\begingroup$ Edit suggestion: Add a definition for $c_{00}$, also it would be better to replace $l_2$ with $l_2(\mathbb{Z})$ either this or add a definition for this too. $\endgroup$ – Yanko Jul 3 '18 at 21:51
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This will solve your problem

Theorem: If $\sum_{n=1}^\infty a_n$ converges then $\lim_{N\rightarrow\infty} \sum_{n=N}^\infty a_n=0$.

The word "This" in the first sentence of this answer is linked to a proof of this Theorem.

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  • $\begingroup$ Thanks for your answer! Does $\lim_{N\rightarrow\infty} \sum_{n=N}^\infty a_n=0$ mean there exists $N_0\in\mathbb{N}$ such that $|\sum_{n=N}^\infty a_n=0|<\epsilon$ for all $N>N_0$ and $\epsilon>0$. $\endgroup$ – Answer Lee Jul 3 '18 at 22:25
  • $\begingroup$ @AnswerLee Yes, that's the definition. Note that it should be $|\sum_{n=N}^\infty a_n-0|<\varepsilon$. The way it's written now makes no sense. $\endgroup$ – Yanko Jul 4 '18 at 10:03
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You have to use that $x\in\ell^2$, so $\sum_{k=1}^\infty | x_k|^2<\infty$.

This implies that $$ \sum_{k=m}^\infty | x_k|^2 \to 0 \quad\text{for}\;m\to\infty. $$

Then you can choose $m\in\mathbb N$ that has the property that you want.

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