Two $p$-norms are not equivalent for different $p$ on $\ell_1$ Given $1\le p < r < \infty$, prove that $\|\cdot\|_p$ and $\|\cdot\|_r$ are not equivalent.
The approach I was trying is as follows:
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Want to show that there do not exist $m,M>0$ such that $$m\|x\|_r\le\|x\|_p\le M\|x\|_r$$
for all sequences $(x_k)\in \ell_1$.
We know that all sequences in $\ell_1$ converge in a $p$-norm, with $p\ge 1$. The idea, I think, is to show that we can always have a convergent sequence whose $p$-norm is bigger than its $r$-norm, regardless of how large our $M$ is. This would imply that there is no upper bound on the right-hand side of the above inequality.
The problem I'm having is how to formalize this. One might think of an upper bound for the right-hand side as follows:

*

*Define a sequence $M_n := \frac{\left(\sum\limits_{k=1}^\infty |x_k|^r \right)^{1/r}}{\left(\sum\limits_{k=1}^\infty |x_k|^p \right)^{1/p}}$, where $n:=\lceil\|x\|_1\rceil$, and then show that $M_n$ diverges.

But I don't think this is a nice way. So I'd appreciate some hints on this.

Let $x_n\in l_1$ where the first $n$ co-ordinates of $x_n$ are each equal to $1$ and the remaining co-ordinates are all $0.$ Let $1\leq p_1<p_2<\infty.$ 
For brevity let $q=(1/p_1+1/p_2)/2$ and $r=(1/p_1-1/p_2)/2 .$ 
Let $y_n=x_n/n^q.$ Then $$\|y_n\|_{p_1}=n^r\geq 1 \; \text { and }\;  \|y_n\|_{p_2} =n^{-r}\to 0 \;\text { as }\; n\to \infty.$$ In the $l_{p_2}$ norm $(y_n)_n$ converges to the vector $0$ but in the $l_{p_1}$ norm $0\not \in Cl(\{y_n\}_n)$ so the topologies are different.
