Prove that, given $q \in [1,\infty] $, then $ l^{p} \hookrightarrow l^{q} $ for all $p \in [1, q]$. Consider the sequence $ x^{(n)}=\bigl( x^{(n)}_{k} \bigr)_{k \in \mathbb{N}_{0} } \ $ defined by \begin{eqnarray*} x^{(n)}_{k} = (n + k)^{−1/3} \ \ \ \ n,k\in \mathbb{N}_{0} \end{eqnarray*} $(i)$ For which $p \in [1,\infty] $ does the sequence $x^{(n)}$ belong to $l^{p}$? Justify the answer.
$(ii)$ Study the convergence of $x^{(n)}$ in $l^{p}$.
I'm a little confused. In the second point do I have to show the convergence with $n$ to $\infty$ or with a fixed $n$ and $k$ to $\infty$ ?
$x^{(n)}\in l^{p}$ iff $\sum_k \frac 1 {(n+k)^{p/3}} <\infty$. This is true iff $p >3$. When $p >3$ we can apply DCT to conclude that $\sum_k \frac 1 {(n+k)^{p/3}} \to 0$ as $n \to \infty$ since the terms are dominated by those of the convergent series $\sum \frac 1 {k^{p/3}}$. Hence $x^{(n)} \to 0$ in the space $l^{p}$ whenever $p >3$.