non-equivalence of $p$-norms in $\ell_1$ 
For $x\in \mathbb{R}^{\mathbb{N}}, $ define $\|x\|_\infty := \sup_{i\in\mathbb{N}}|x_i|$ and for $1\leq p < \infty,$ the $p$-norm $\|x\|_p$ is given by $\|x\|_p := (\sum_{i} |x_i|^p)^{1/p}.$ The $\ell_p$ space is the set of all sequences with real entries with finite $p$-norm. Two norms $\|\cdot \|_a$ and $\|\cdot \|_b$ are equivalent on a real vector space $V$ if there are $m, M > 0$ so that $m\|x\|_a \leq \|x\|_b \leq M\|x\|_a$ for all $x\in V.$
Show that for $1\leq p < r \leq \infty,$ the norms $\|\cdot \|_p$ and $\|\cdot \|_r$ on $\ell_1$ are not equivalent.

I know how to show that $\ell_1\subseteq \ell_p$ for each $1\leq p\leq \infty.$ I think I should deal with the case where $r=\infty$ first as that might be easier. Taking the negation of the definition of equivalence, I think I should show that $\forall m, M > 0,$ there exists $x \in \ell_1$ so that $m\|x\|_p > \|x\|_r$ or $\|x\|_r > M\|x\|_p.$ However, I'm not sure how to find this particular $x$ for each choice of $m$ and $M.$

Consider the all-ones vector $u^{(n)}$ on $n$ elements. For $1\leq p < r < \infty$, we have
$$\begin{align}
\|u^{(n)}\|_1 &= n
\\
\|u^{(n)}\|_p &= n^{1/p}
\\
\|u^{(n)}\|_r &= n^{1/r} 
\\
\|u^{(n)}\|_\infty &= 1 
\end{align}$$
The first shows that, indeed, $u^{(n)}\in\ell_1$ for all $n\geq 1$. The last 3 show what you want: those norms cannot be equivalent, as the ratios can be made arbitrarily large/small as $n\to\infty$.

If you want some intuition: this is basically the type of vectors that makes Holder's inequality tight (worst possible correspondence between $\ell_p$ norms).

Also, as an important comment: this will show that for $1\leq p<r\leq \infty$, the norms are not equivalent, by showing there cannot be a constant $M>0$ (depending only on $p$ and $r$) such that
$$
\|\cdot\|_p \leq M\|\cdot\|_r
$$
However, the other inequality will always hold with $m=1$: $\|\cdot\|_r \leq \|\cdot\|_p$. This follows from the monotonicity of $\ell_p$ norms.
