Find a sequence converging to zero but not the element of $\ell^p$ space for any $1

Question

I am studying functional analysis and I have a problem about finding a sequence converging to zero such that this sequence is not in $\ell^p$ for any $p$. By $\ell^p$, I mean

$$\ell^p := \left\{ (x_k)=(x_1,x_2,...):\sum_{k=1}^\infty|x_k|^p<\infty \right\}$$ where $1<p<\infty$.

First, I thought of the simple sequence $(1/k)_{k \in \mathbb{N}}$ which converges to zero, but, then, I realized it is an element of $\ell^p$ when $p>2$. I thought of a couple more examples, but they did not work either. Can somebody help me out here?

Answer 1

Try the sequence $x_k = 1/\ln(k+1)$.

Answer 2

If $k \in \mathbb N_{\geq 2}$ $$|\ln(k+1)| > 1$$ Now if $p \in \mathbb N_{\geq 1}$ $$|\ln(k+1)|^p \gt |\ln(k+1)| \implies \left| \frac{1}{\ln(k+1)} \right|>\left|\frac{1}{\ln(k+1)}\right|^p$$ By the Raabe-Duhamel's Test, consider the sequence, $$b_n=n \left(\frac{a_n}{a_{n+1}}-1\right)$$ where $$a_n=\frac{1}{ln(1+n)^p}$$ Then $$ \begin{array}\\ L=\lim\limits_{n \to \infty} b_n\\ =\lim\limits_{n \to \infty} n\left(\frac{\ln(2+n)^p}{\ln(1+n)^p}-1\right)\\ =\lim\limits_{n \to \infty} \frac{n}{\ln(n+1)}\ln\left(\frac{2+n}{1+n}\right)\sum_{i=0}^{p-1} (-1)^{p-i-1}\left( \frac{\ln(2+n)}{\ln(1+n)} \right)^i\\ =(-1)^{p-1}\lim\limits_{n \to \infty} \frac{n}{\ln(n+1)}\ln\left(\frac{2+n}{1+n}\right)\\ =(-1)^p \lim\limits_{t \to 0+} \frac{\ln(1+t)}{t\ln(t)} \qquad\text{(Changing Variables } t=\frac{1}{n+1})\\ =(-1)^p \lim\limits_{t \to 0+} \frac{1}{(1+t)(1+\ln(t))}\qquad\text{(By L' Hopital's Rule)}\\ = 0 < 1\\ \end{array} $$

Hence, the series $\sum_{n\geq1} a_n$ diverges for all $p \geq 1$ i.e. the sequence $\left(\frac{1}{ln(n+1)}\right)_{n\geq1}$ is not $p$-th summable for any $p\geq1$ but converges to $0$.

Answer 3

@Robert Israel's answer is great. And I think the easiest way to prove that the sequence $(x_i), x_i = 1 / \ln (i+1)$ is not in $\ell^p$, where $p$ is a real number and $1 \leq p < \infty$, is using Schlömilch's Generalization of Cauchy Condensation Test.

This generalization establishes:

$$ \sum_{n=1}^{\infty}{f(n}) \, \mbox{ converges if and only if } \, \sum_{n=0}^{\infty}{(u(n+1)-u(n)) f(u(n))} \mbox{ converges, } $$ where $u(n)$ is a strictly increasing sequence of positive integers such that there is a positive real number $N$, for which:

$$ \frac{u(n+1)-u(n)}{u(n)-u(n-1)} \, < \, N \mbox{ para todo } n. $$

Setting $u(n) = 2^n-1$ and then we have

$$ \sum_{i=1}^{\infty} {\frac{1}{\ln (i+1) ^p}} \, \mbox{ converges if and only if } \, \sum_{i=0}^{\infty}{ \frac{2^i}{(\ln (2^i) ) ^p }} = \frac{1}{(\ln (2))^p} \, \sum_{i=0}^{\infty}{\frac{2^i}{i^p}} \, \mbox { converges } $$

Since, for all $p$, the term $(\ln(2))^p$ is a constant, we only have to investigate the convergence of the last series.

We will use the ratio test, where, if $L$ is the limit of the ratio of two consecutive terms of the sequence, then, if $L< 1$, the series converges, and, if $L> 1$, the series diverges.

$$ L = \lim_{n\to\infty} \frac{2^{n+1}}{2^n} \cdot \frac{(n+1)^p}{n^p} = 2 \cdot \lim_{n\to\infty} \left( \frac{n+1}{n} \right)^p = 2 $$

Since $L>1$, the series diverges, as does the original series.