Prove this inclusion: $\bigcup_{k<p}\ell^k\subsetneq\ell^p$

Question

Let $1<p<\infty$. I have to prove that $$ \bigcup_{k<p}\ell^k\subsetneq\ell^p. $$ I am not able to find a counterexample to prove the inequality.

Answer 1

Take $x(n):=n^{-1/p}\log(n)^{-2/p}$, $n\geqslant 2$.

Since $|x(n)|^p=n^{-1}\log(n)^{-2}$, $x$ belongs to $\ell^p$. For $q\lt p$, we can show that $|x(n)|^q\geqslant n^{-\beta}$ for some $\beta\lt 1$ and each $n$ large enough. Indeed, $$|x(n)|^q=\frac 1{n^{q/p}(\log n)^{2q/p}}.$$ Now take $\beta\in (q/p,1)$. Then $$|x(n)|^q=\frac 1{n^\beta}\cdot \frac{n^{\beta-q/p}}{(\log n)^{2q/p}}.$$ Since $\frac{n^{\beta-q/p}}{(\log n)^{2q/p}}\to \infty$, we have for $n$ large enough $\frac{n^{\beta-q/p}}{(\log n)^{2q/p}}\geqslant 1$.