About the $p$ summable sequences How to prove that $\ell^p \setminus \cup_{1\leq q <p}\ell^q$ is not empty?
Here $\ell^p=L^p(\mathbb{N})$.
 A: If $p=+\infty$, just take $x(n)=1$ for all $n\in \mathbb{N}$. If $p=1$, there is nothing to prove. If $p\in(1,+\infty)$ consider
$$
x(n)=\frac{1}{(n+2)^{1/p}\log(n+2)}
$$
Thus we have
$$
\Vert x\Vert_p=
\left(\sum\limits_{n=1}^\infty\frac{1}{(n+2)\log^p(n+2)}\right)^{1/p}\leq
\left(\int\limits_{0}^\infty\frac{du}{(u+2)\log^p(u+2)}\right)^{1/p}=
$$
$$
\left(\int\limits_{2}^\infty\frac{du}{u\log^p u}\right)^{1/p}=
\left(\int\limits_{2}^\infty\frac{du}{u\log^p u}\right)^{1/p}=
\left((1-p)^{-1}\log^{1-p} 2\right)^{1/p}<
+\infty
$$
So, $x\in l^p$. On the other hand for all $q\in[1,p)$ we have
$$
\Vert x\Vert_q=
\left(\sum\limits_{n=1}^\infty\frac{1}{(n+2)^{q/p}\log^q(n+2)}\right)^{1/q}\geq
\left(\int\limits_{1}^\infty\frac{du}{(u+2)^{q/p}\log^q(u+2)}\right)^{1/q}=
\left(\int\limits_{3}^\infty\frac{du}{u^{q/p}\log^q u}\right)^{1/q}
$$
Now,I will prove that the last integral is diverges. Since $p>q$, then 
$$
\lim\limits_{u\to\infty}\frac{u^{\frac{p-q}{2p}}}{\log^q u}=+\infty
$$
and as the consequence there exist $u_0>3$ such that $$\frac{u^{\frac{p-q}{2p}}}{\log^q u}>1
$$
Then
$$
\int\limits_{u_0}^\infty\frac{du}{u^{q/p}\log^q u}=
\int\limits_{u_0}^\infty \frac{u^{\frac{p-q}{2p}}}{\log^q u}\frac{du}{u^{1-\frac{p-q}{2p}}}>\int\limits_{u_0}^\infty \frac{du}{u^{1-\frac{p-q}{2p}}}=+\infty
$$
$$
\Vert x\Vert_q\geq
\left(\int\limits_{3}^\infty\frac{du}{u^{q/p}\log^q u}\right)^{1/q}>
\left(\int\limits_{u_0}^\infty\frac{du}{u^{q/p}\log^q u}\right)^{1/q}=+\infty
$$
So $x\notin l^q$ for all $q\in[1,p)$. Hence $x\in l^p\setminus\left(\bigcup\limits_{1\leq q<p} l^q\right)$
A: Other users gave an explicit example. Baire theorem, as suggested by Jose27, can be also used. We argue by contradiction. Define
$$F_n:=\left\{x\in\ell^p\mid \sum_{k=1}^\infty|x_k|^{p-\frac 1n}\leqslant n\right\}.$$
Since convergence in $\ell^p$ implies pointwise convergence, each $F_n$ is closed and $\ell^p=\bigcup_nF_n$ (under the assumption $\bigcup_{1\leqslant q\lt p}\ell^q=\ell^p$) because if $x\in \ell^{p-\frac 1{n_0}}$, then $\lVert x\rVert_{p-\frac 1n}\leqslant\lVert x\rVert_{p-\frac 1{n_0}}$ for $n\geqslant n_0$ so take $n$ such that $n^{p-\frac 1n}\geqslant\lVert x\rVert_{p-\frac 1{n_0}}$.
By Baire categories theorem, we would get that $\ell^p$ is contained in some $\ell^{p_0}$ for some $p_0\lt p$, which is not possible. 
A: chessmath, are you related to Some Guy?
If not, see question Is there a null sequence that is in not in $\ell_p$ for any $p<\infty$? and use the fact in my answer there.
A: Hint: Try to construct an element of the form $(n^\alpha)_{n\in\mathbb{N}}$ for a well chosen $\alpha\in\mathbb{R}$.
