Is there a null sequence that is in not in $\ell_p$ for any $p<\infty$? Is 
$$\bigcup_{p<\infty}\ell_p=c_0 ?$$
At least one inclusion obvious: every $p$-summable sequence converges to zero.
 A: Consider
$$
x(n)=\frac{1}{\log(n+1)}
$$
It is easy to check that $x\in c_0\setminus \left(\bigcup_{1\leq p<\infty}\ell_p\right)$. Indeed
$$
\lim\limits_{n\to \infty} x(n) = \lim\limits_{n\to \infty}\frac{1}{\log (n+1)}=0
$$
so $x\in c_0$. Now for fixed $p\in [1,+\infty)$ there exist $N\in \mathbb{N}$ such that for all $n>N$ we have $\log^p (n+1)<n$, so
$$
\Vert x\Vert_p=
\left(\sum\limits_{n=1}^\infty \frac{1}{\log^p (n+1)}\right)^{1/p}=
\left(\sum\limits_{n=1}^N \frac{1}{\log^p (n+1)}+
\sum\limits_{n=N+1}^\infty \frac{1}{\log^p (n+1)}\right)^{1/p}>
$$
$$
\left(\sum\limits_{n=1}^N \frac{1}{\log^p (n+1)}+
\sum\limits_{n=N+1}^\infty \frac{1}{n}\right)^{1/p}=+\infty
$$
the last equality holds since the series
$$
\sum\limits_{n=N+1}^\infty \frac{1}{n}
$$
diverges.
A: Prove: If $V$ is a separable Banach space, then $V$ cannot be written as a countable union of proper linear subspaces.  
Because of inclusions among the $\ell_p$ spaces, your union is, in fact, equal to a countable union.
A: $\frac{1}{\ln n}$ converges to 0 but $\sum \frac{1}{\ln^p n} = \infty$ for any $p \geq 1$.
A: $$
\lim_{n\to\infty}\frac{1}{\log(n+1)}=0\text{, but }\sum_{n=1}^{\infty}\frac{1}{(\log(n+1))^p}=\infty\quad\forall p>0.
$$
