Ambiguous definition of a sequence in $l^2$

Question

Show that the sequence $(x_n)_{n\in\mathbb{N}}$ in $l^2$ given by $$ (x_n)_{n\in\mathbb{N}} := \left(1,\frac{1}{2},\frac{1}{3},...,\frac{1}{n},0,0,0,...\right) $$ converges to $$x=\left(1, \frac{1}{2},\frac{1}{3},...,\frac{1}{n},\frac{1}{n+1},...\right)$$ in $l^2$.

Firstly I'm having a problem to understand the first definition. It defines a class of sequences of $l^2$ in the RHS, not just one sequence as stated in the LHS. And to say that this sequence converges to a sequence in $l^2$ it must be the case that it's actually a sequence $((x_n)_{n\in\mathbb{N}})_{n\in\mathbb{N}}$ of sequences of $l^2$. Seems alright?

That being right, I still don't know how to calculate the limit. Given $\varepsilon > 0$, I must find an $N$ such that $n>N$ implies that the sequence $(1,\frac{1}{2},...,\frac{1}{n},0,0,0...)$ is $\varepsilon$-distant to $(1,\frac{1}{2},\frac{1}{3},...,\frac{1}{n},\frac{1}{n+1},...)$. But $$\|(1,\frac{1}{2},...,\frac{1}{n},0,0,0...)-(1,\frac{1}{2},\frac{1}{3},...,\frac{1}{n},\frac{1}{n+1},...)\| = \|(0,0,0...,\frac{1}{n+1},\frac{1}{n+2},...)\| = $$ $$ \sum_{k=n+1}^{\infty} \frac{1}{k^2}, $$ and I don't know what to do with this sum. Any clarification will be appreciated.

Answer 1

Your understanding of the first equality is correct: the right-hand side should have been a sequence of elements in $l^2$, instead of a single element of $l^2$. Your notation is incorrect, however: $x_n$ represents a single element of $l^2$ (namely $(1,1/2, \ldots, 1/n, 0, 0, \ldots)$), so the sequence of elements of $l^2$ is $(x_n)_{n \in \mathbb{N}}$. Each $x_n$ is itself a sequence.

For the second part, your work so far is good. Just note that $\sum_{k \ge 1} \frac{1}{k^2}$ is a convergent sequence, so the tail sum $\sum_{k = n+1}^\infty \frac{1}{k^2}$ decreases to zero as $n \to \infty$.

[Finding an explicit expression for $N$ in terms of $\epsilon$ is not really necessary, but if you must, you can bound the tail sum by $\int_n^\infty x^{-2} \, dx$ or something like that.]

Answer 2

For each natural number $n$, $x_n$ is the sequence$$\left(1,\frac12,\frac13,\ldots,\frac1n,0,0,\ldots\right).$$So, each $x_n$ is a sequence, and it is clear that each $x_n$ belongs to $\ell^2$. So, the sequence $(x_n)_{n\in\mathbb N}$ may converge to$$\left(1,\frac12,\frac13,\ldots,\frac1n,\frac1{n+1},\ldots\right),\tag1$$which is also an element of $\ell^2$. But does it converge? Yes, because$$\bigl\|(1)-x_n\bigr\|_2=\sqrt{\sum_{k=n+1}^\infty\frac1{k^2}}$$and, since the series $\displaystyle\sum_{k=1}^\infty\frac1{k^2}$ converges (to $\dfrac{\pi^2}6$, but that's not important here), we have that $\displaystyle\lim_{n\to\infty}\sum_{k=n+1}^\infty\frac1{k^2}=0$.