Is the union of $l^p$ spaces ($1\le p<\infty$) equal to $c_0$?

Question

Is it true that $$\bigcup_{p\geq1} l^p=c_0?$$

I have shown that since $\|x\|_p$ is finite for any $x\in l^p, 1\leq p<\infty$, if the sequence $x=(x_1, x_2, x_3, \dots)$ does not converge to $0$ then it converges to some other constant since the sum is finite. Then the sum will become infinite which is a contradiction. So every element of $l^p$ is in $c_0$. Now this answer has me confused as this implies that we can't prove the converse inclusion but I clearly remember that in class I was taught that it is equal. So which is true?

Answer 1

$\frac{1}{\log (2+n)}\in c_0$. But since for $x\gg 1$, $\log x\le C_q x^q$ for any $q> 0$, in particular $(\log x)^p \le C_p x$ for all $p\in[1,\infty)$, so $$\frac1{(\log (2+n))^p} \ge \frac{C_p}{2+n}$$
which is not summable.