Prove $\ell^5$ is contained in $\ell^6$. I am struggling with the proof to show that, for any $p$, $r$ such that  $1 \le p <r < \infty$, that $\ell^ p\subset\ell ^r$.
Could somebody please give a helpful nudge by showing how this works for a specific case, for example, prove that $\ell^5 \subset\ell^6$ ?
Many thanks.
 A: Let $p,q$ such as $1\leq p<q$ and $x = \{x_k\} \in l^p$. Then by definition $\sum_{k=1}^{\infty} {|x_k|^p} < \infty$.
Therefore $|x_k| \to 0$ as $k \to \infty$.
This implies that there exists an index $k_0$ such that $|x_k| < 1$ for any $k ≥ k_0$.
So for $k ≥ k_0$ we have $|x_k|^q < |x_k|^p$ which implies $\sum_{k=1}^{\infty} {|x_k|^q} < \infty$  thanks to the comparison test for series.
Thus $x \in l^q$ and therefore $l^p\subset l^q$

Now, that is the proof. In order to help you grasp it, as you ask, lets say $p=5, q=10$ and $x = \{ \frac{1}{n^{1/20}}\}$ .
Lets see what happens: $\sum_{n=1}^{\infty}{|x_k|^p} = \sum_{n=1}^{\infty}{|\frac{1}{n^{1/20}}|^5} = \sum_{n=1}^{\infty}{|\frac{1}{n^{1/4}}|} $
And $\sum_{n=1}^{\infty}{|x_k|^q} = \sum_{n=1}^{\infty}{|\frac{1}{n^{1/20}}|^{10}} = \sum_{n=1}^{\infty}{|\frac{1}{n^{1/2}}|} $
Now, we can see that from a point on (in this case $n=2$), $\frac{1}{n}<1$ and $\frac{1}{n}^{1/2}<\frac{1}{n}^{1/4}$  for all $n\geq 2$. That means that the second series is less than, or equal to the first. And if the first one converges, then so does the second one.
A: Hint: use the Holder inequality.
