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To show that $M=\{x=(x_1,x_2,...)\in \ell_2:\sum_{n=1}^{\infty}{1\over n}x_i=0\}$, a solution suggest that $M=\{x\in \ell_2:\langle x,({1\over n})_n\rangle =0\}$ and by continuity of inner product, for $x^{(m)}\to x$ as $m\to \infty$, $\langle x,({1\over n})_n\rangle=0$.

But for $M=\{x=(x_1,x_2,...)\in \ell_2:\sum_{n=1}^{\infty}{1\over \sqrt{n}}x_i=0\}$, the solutions use a pretty long disproof. When the Inner Product continuity claim was used, it felt like I was told good news and wouldn't have to worry about ${1\over n}$ itself, but as soon as I found out this is not the case for ${1\over \sqrt{n}}$, it made me wonder what enabled the Authors to immediately ignore the $({1\over n})$ sequence and go right to the inner product property. What sets the difference between the two sequences used in any $M$ set? It didn't even seem as though ${1\over n}$ has to be compared to elements of $l_2$.

Could you give me a hint?

Notations: $\ell_2$ contains infinite sequences $(x_1,x_2,...)$ such that $|\langle x,x\rangle|=({\sum_{i=1}^{\infty}|x_i|^2})^{1\over 2}<\infty$.

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    Maybe the fact that $(1/n)_n\in\ell_2$, but $(1/\sqrt{n})_n\notin\ell_2$? (so the inner product argument makes sense for the former, but not the latter -- the inner product with $(1/\sqrt{n})_n$ is not defined) – 2017-02-27
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    I just notices that in the norm, the sum goes over $x_i^2$'s and not $x_i$'s. It is my bad, I guess. – 2017-02-27
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    I mean, I knew that, but it slipped off my mind. – 2017-02-27

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