Is $\ell^p$ a subset of $c_0$ for all finite $p$?

Question

Sorry if this is a too straightforward question, but I just want to verify that my reasoning is correct.

If a sequence $x$ is in $\ell^p$ then $$\|x\|_p^p=\sum\limits_{i=1}^{\infty} |x_i|^p < \infty$$ which implies in particular that $\lim\limits_{n \to \infty} |x_n|^p = 0$, and also that $\lim\limits_{n \to \infty} |x_n| = 0$ and hence that $\lim\limits_{n \to \infty} x_n = 0$, which means that $x \in c_0$ as well. So $\ell^p \subset c_0$.

The other inclusion does not hold however, as the sequence $\{\frac{1}{n}\} \in c_0$ but it is well known that $\sum\limits_{n=1}^{\infty} \frac{1}{n}$ does not converge. What do you think?

Answer 1

Your argument to show $\ell^p \subset c_0$ is correct.

Your argument to show the opposite inclusion does not hold only works for $\ell^1$. But it can easily be extended to show that $c_0 \not\subset \ell^p$ by noting that $\left( \frac{1}{\sqrt[p]{n}} \right) \in c_0$, but its $p$-th power is not summable, so that $\left( \frac{1}{\sqrt[p]{n}} \right) \not\in \ell^p$.

Answer 2

Well $\sum\frac{1}{n^2} <\infty$, so $(1/n)_{n\in\Bbb N}\in \ell^2$. But you can argue $$ \Big( \frac{1}{n^{1/p}} \Big) \in c_0 \setminus \ell^p. $$