Convergent sequences If $l^p=\{\langle x_k \rangle \in R^n | \sum_{k=1}^{\infty} |x_k|^p < \infty \}$ and $1\leq p<q$, then $l^p\subset l^q$.
Prove.  
I know that I have to show that if $x=\langle x_k \rangle\in l^p\rightarrow x\in l^q$ and I can use a fact that $l^p\subset c_0$, where $c_0=\{x=\langle x_k \rangle \in R^n | \displaystyle \lim_{k \to +\infty} x_k=0\}$.
 A: One hint is that for a number $x$ less than 1, $x^q$ is less than $x^p$. And the numbers in the sequence are approaching 0.
A: Suppose $1\le p<q\le\infty.$ We will prove that $\ell^{p}\subseteq \ell^{q}$ by proving that $\|a\|_{q}\le\|a\|_{p}$ for any $a\in \ell^{p}.$
Define $b_n := \frac{a_n}{\|a\|_{p}}$ and observe that $\vert b_n \vert \leq 1$ for all $n \in \mathbb N$. Notice the following chain of equivalent statements:
                    \begin{align*}
      \vert b_n \vert^{q} &\leq \vert b_n \vert^{p} \\
      \sum_{n = 1}^\infty \vert b_n \vert^{q} & \leq \sum_{n = 1}^\infty \vert b_n \vert^{p} \\
      \sum_{n = 1}^\infty \bigg \vert \frac{a_n}{\|a\|_{p}} \bigg \vert^{q} & \leq \sum_{n = 1}^\infty \bigg \vert \frac{a_n}{\|a\|_{p}} \bigg \vert^{p} \\
      \Bigg(\sum_{n = 1}^\infty \bigg \vert \frac{a_n}{\|a\|_{p}} \bigg \vert^{q}\Bigg)^{1/p} & \leq \Bigg( \sum_{n = 1}^\infty \bigg \vert \frac{a_n}{\|a\|_{p}} \bigg \vert^{p} \Bigg)^{1/p}\\
      \frac{1}{\big(\|a\|_{p}\big)^{q/p}} \Bigg(\sum_{n = 1}^\infty \vert a_n \vert^{q} \Bigg)^{1/p} & \leq 1\\
      \Bigg(\sum_{n = 1}^\infty \vert a_n \vert^{q} \Bigg)^{1/p} & \leq \big(\|a\|_{p}\big)^{q/p}\\
      \sum_{n = 1}^\infty \vert a_n \vert^{q}  & \leq \big(\|a\|_{p}\big)^{q}\\
      \Bigg(\sum_{n = 1}^\infty \vert a_n \vert^{q} \Bigg)^{1/q} & \leq \|a\|_{p}\\
       \|a\|_{q}& \leq \|a\|_{p}.
     \end{align*}
You didn't ask for the case when $q = \infty$, but it holds there too. Notice that $\vert a_n \vert \leq \bigg(\sum_{n = 1}^\infty \vert a_n \vert^{p} \bigg)^{1/p}$ for all $n \in \mathbb N$, and thus $$\|a_n\|_{q} = \|a_n\|_\infty = \sup_{n \in \mathbb N} \vert a_n \vert \leq \bigg(\sum_{n = 1}^\infty \vert a_n \vert^{p} \bigg)^{1/p} = \|a_n\|_{p}.$$
