For $p>1$ show that there exists a sequence $A=(a_n)_{n\in\mathbb{N}}$ of reals numbers such that $||A||_p <\infty$ while $\sup_{n\in\mathbb{N}}(\sum_{i=n}^{2n}a_i)=\infty$.
I thought $a_n=1/n$ because of the first condition but the second seems to be finite.. I can't see how to make its convergence to $0$ slower without breaking the first condition. Is it possible?!
Let $(c_N)_N$ be a sequence of non-negative numbers that will be specified later. If $2^N\leqslant i\lt 2^{N+1}-1$, define $a_i$ as $c_N$. Then $$\sup_{n\in\mathbb{N}}\sum_{i=n}^{2n}a_i\geqslant \sup_{N\geqslant 1}\sum_{i=2^N}^{2^{N+1}-1}a_i=\sup_{N\geqslant 1}2^Nc_N $$ and $$ \sum_i a_i^p=\sum_N\sum_{i=2^N}^{2^{N+1}-1}a_i^p=\sum_N2^Nc_N^p. $$ The choice $c_N:=N2^{-N}$ works.