Sequence in $l^p$ but not $l^q$ for all $q<p$

Question

I need to find a sequence for real $p>1$ so it is in $l^p$ but not in any of the space $l^q$ with $1 \leq q <p$.

I tried the sequence $(1/n)^{1/q}$ which is in $l^p\setminus l^q$. However, this depends on $q$. I need one sequence that works for all $q<p$.

Answer 1

Hint: Think about

$$\sum_{n=2}^{\infty}\frac{1}{n^a (\ln)^b}$$

for $a, b> 0.$ This diverges for all $a, 0< a <1,$ but converges for $a= 1,b>1.$

Answer 2

Consider that if $a_n<p$ for each $n$ with $\lim_{n\to \infty}a_n=0$, and $x=(x_n)_{n\in N}=(n^{-1/(p-a_n)})_{n\in N}$ then $x\notin l_q$ for any $q<p.$ I tried putting $(x_n)^p=n^{-1}(\log n)^{-2}$ for all sufficienly large $n$, to get $x\in lp.$ For $n>1$, I got $a_n=p y_n/(1+y_n)$ where $y_n=2(\log (\log n))/\log n.$ And $n\geq 4\implies a_n<p.$ So for $1<p<\infty$ we have $\cup_{1\leq q<p}l_q\ne l_p.$