Let $s=\{1/k\}^{\infty}_{k=1}$. Find a sequence $\{s_n\}^{\infty}_{n=1}$ of points in $l^2$ such that each $s_n$ is distinct from $s$ and such that $\{s_n\}^{\infty}_{n=1}$ converges to $s$ in $l^2$.
This is a problem from Goldberg. Exercise 4.3, 3. I don't understand the question. How can I find a sequence that converges to another sequence? Is $s$ the limit of $\{1/k\}$ or it is the variable assigned to the sequence itself? If it is the limit then it can be worked out
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@RobertZ was much quicker than me, but here is another example in case you need it. Put $$ s_n = \left\{\frac{1}{k+\frac{1}{n}}\right\}_{k=1}^\infty. $$ Then $s_n \in l^2$ for each $n \geq 1$ by comparison with $\sum 1/k^2,$ i.e. $$ \sum_{k=1}^\infty \left(\frac{1}{k+\frac{1}{n}}\right)^2 \leq \sum_{k=1}^\infty \frac{1}{k^2} < \infty, $$ and $s_n \to s$ in $l^2$ since $$ \sum_{k=1}^\infty \left( \frac{1}{k+\frac{1}{n}}-\frac{1}{k} \right)^2 = \frac{1}{n^2}\sum_{k=1}^\infty \left(\frac{1}{\left(k+\frac{1}{n}\right)k}\right)^2 \leq \frac{1}{n^2}\sum_{k=1}^\infty \frac{1}{k^4} \to 0, \text{ as } n\to \infty, $$ since the last sum is bounded.
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Consider $s_n = \left(1, \frac12, \ldots, \frac1n, 0, 0, \ldots\right) \in \ell^2$. We have
$$\|s - s_n\|_2^2 = \left\|\left(0, , \ldots, 0, \frac1{n+1}, \frac1{n+2}, \ldots\right)\right\|_2^2 = \sum_{k=n+1}^\infty \frac1{k^2} \xrightarrow{n\to\infty} 0$$
because $\sum \frac1{k^2}$ is a convergent series. Therefore $(s_n)_{n=1}^\infty$ converges to $s$ in $\ell^2$.
Hint. Recall that $l^2$ is the normed space of all square summable sequences $s$, such that $$\|s\|^2_2=\sum_{k=1}^{\infty}s(k)^2<+\infty.$$ Let $s_n(k):=\frac{1}{k}+\frac{1}{nk}$ (also $s_n(k):=\frac{a_n}{k}$ with $a_n\to 1$ will work), then $\{s_n(k)\}_{k\geq1}\in l^2$ for each $n\geq 1$. Now consider the limit of $$\lim_{n\to \infty}\|s_n-s\|^2_2=\lim_{n\to \infty}\sum_{k=1}^{\infty}\left(s_n(k)-\frac{1}{k}\right)^2$$ and show that it is zero.