Vtyjkyuk

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

Hint. Recall that $l^2$ is the normed space of all square summable sequences $s$, such that
$$|s|^2_2=sum_k=1^inftys(k)^2<+infty.$$
Let $s_n(k):=frac1k+frac1nk$ (also $s_n(k):=fraca_nk$ with $a_nto 1$ will work), then $s_n(k)_kgeq1in l^2$ for each $ngeq 1$. Now consider the limit of
$$lim_nto infty|s_n-s|^2_2=lim_nto inftysum_k=1^inftyleft(s_n(k)-frac1kright)^2$$
and show that it is zero.

Consider $s_n = left(1, frac12, ldots, frac1n, 0, 0, ldotsright) in ell^2$. We have

$$|s - s_n|_2^2 = left|left(0, , ldots, 0, frac1n+1, frac1n+2, ldotsright)right|_2^2 = sum_k=n+1^infty frac1k^2 xrightarrowntoinfty 0$$

because $sum frac1k^2$ is a convergent series. Therefore $(s_n)_n=1^infty$ converges to $s$ in $ell^2$.

@RobertZ was much quicker than me, but here is another example in case you need it. Put
$$ s_n = leftfrac1k+frac1nright_k=1^infty. $$
Then $s_n in l^2$ for each $n geq 1$ by comparison with $sum 1/k^2,$ i.e.
$$ sum_k=1^infty left(frac1k+frac1nright)^2 leq sum_k=1^infty frac1k^2 < infty, $$
and $s_n to s$ in $l^2$ since
$$ sum_k=1^infty left( frac1k+frac1n-frac1k right)^2
= frac1n^2sum_k=1^infty left(frac1left(k+frac1nright)kright)^2
leq frac1n^2sum_k=1^infty frac1k^4 to 0, text as nto infty, $$
since the last sum is bounded.