Let $x^{(n)}$ be a the sequence in $\mathscr{l^2}(\mathbb{R})$ defined by $x_k^{(n)}:=\frac{1}{n+k}$. Show that the sequence $x^{(n)}$ converges to zero in $\mathscr{l^2}(\mathbb{R})$.
Attempt:
I showed for every $k$, the $k^{th}$ term tends to zero as $n\rightarrow{\infty}$. However this is saying the sequence converges in a 'point wise' sense to the zero sequence and I suspect this is not enough.
I used the definition of convergence using the $\|.\|_{l^2}$ norm but struggled to complete this.
Additional question of notation: why does $\mathscr{l}^p$ denote sequence spaces? What does the small cursive $l$ stand for?
Not sure about the $l$ notation, but the answer is simple, and you have done some of the groundwork : the definition tells you that you must prove that $\| x_k^{(n)}\|_2 \to 0$ as $n \to \infty$. Therefore, you must prove that $\displaystyle\lim_{n \to \infty} \sum_{k=1}^{\infty} \frac 1{(n+k)^2} = 0$ (this implies the square root of LHS also converges to $0$, as desired).
Now, you have proved that for each $k$, $x_{k}^{(n)} \to 0$. Imagine you could exchange the summations : $$ \boxed{\lim_{n \to \infty} \sum_{k=1}^\infty \frac 1{(n+k)^2} = \sum_{k=1}^\infty \lim_{n \to \infty} \frac 1{(n+k)^2}} = \sum_{k=1}^\infty 0 = 0 $$
then we'd be done.
But what lets us exchange this? That is the dominated convergence theorem : as long as you are able to dominate all $x_{k}^{(n)}$ pointwise by a sequence $y_k$ such that $\|y_k\| < \infty$, then the box is exactly the conclusion of DCT. But then, the good news is that $x_{k}^{(n)}$ is decreasing in $n$, so $x_k^{(1)}$ itself works!
Finally the conclusion is clear.