Prove $\|\cdot\|_p$ and $\|\cdot \|_q$ aren't equivalent on $\ell^p$ $1 \leq p < q < \infty$
So we need to find something in $\ell ^p$ that gives different results in each of the two norms but I can't think of anything. 
I could think of something that is n $\ell ^q$ and then show that they are not equivalent but nothing in $\ell ^p$
 A: Take the sequence (of sequences) $(x^{(n)})_n\in \left(\mathbb{R}^{\mathbb{N}}\right)^{\mathbb{N}}$ defined by $x^{(n)}_k =\frac{1}{k^{\frac{1}{p}+\frac{1}{n}}}$. 


*

*For any $n$, you have $$\lVert x^{(n)} \rVert_p = \left( \sum_{k=1}^\infty \frac{1}{k^{1+\frac{p}{n}}} \right)^{1/p}< \infty$$
and
$$\lVert x^{(n)} \rVert_q = \left( \sum_{k=1}^\infty \frac{1}{k^{\frac{q}{p}+\frac{q}{n}}} \right)^{1/q}< \infty$$ (since $\frac{q}{p} > 1$) so $(x^{(n)})_n\in \ell_p^{\mathbb{N}}$ and $(x^{(n)})_n\in \ell_q^{\mathbb{N}}$.

*However, you can check that $(x^{(n)})_n$ converges in $\ell_q$, but not in $\ell_p$.${}^{(\dagger)}$ So the two norms cannot be equivalent, since otherwise $(x^{(n)})_n$ would have the same nature under both norms $\lVert\cdot\rVert_q$ and $\lVert\cdot\rVert_p$.
$(\dagger)$ This is because, where $n\to \infty$, the limit of $(x^{(n)})_n$ can only be the pointwise limit $x\in\mathbb{R}^\mathbb{N}$ defined by $x_k = \frac{1}{k^{\frac{1}{p}}}$. But $x\in\ell^q$, yet not in $x\in\ell^p$.
