Sequences not in $l^p$

Question

I am wondering if there is an easy sequence $x_n \in \mathbb R$ with $x_n \to 0$ and $x_n \notin l^p$ for all $1 \le p < \infty$.

I found $x_n = (\log n)^{-1}$ satisfies $x_n \to 0$ and $x_n \notin l^p$ because

$\sum_{n=2}^\infty |x_n|^p \ge \sum_{n=2}^\infty |x_{n+1}|^p \ge \int_2^\infty (\log x)^{-p} dx \ge \int_2^\infty (\log x)^{-1} dx = (x(\log x -1))_2^\infty = \infty$.

But it is complicated. Is there an easier example?

I found one: $x_n = n^{1/n}$. But the proof I wrote in the question for the other sequence is incorrect. — mathvisitor, Feb 13 '13 at 13:56
$n^{1/n} \rightarrow 1$, not zero. (Take log, apply L'Hospital's rule.) — abnry, Feb 13 '13 at 13:57
@DavidMitra Thank you. I will have to search further for better example. — mathvisitor, Feb 13 '13 at 13:58
@DavidMitra Thank you for help. I was hoping to find easier example without a logarithm. — mathvisitor, Feb 13 '13 at 14:10
For each integer $m$, select $m_n$ so that $(\underbrace{1/m,1/m,\ldots,1/m}_{m_n-\text{terms}})$ has $\ell_m$ norm exceeding $1$. Then paste these sequences together. (I don't think this is any easier than using your sequence.) — David Mitra, Feb 13 '13 at 14:13

Julien · Accepted Answer · 2013-02-13T14:18:26.777

1

Yes, your example is good.

A Bertrand series $$ \sum_{n\geq 2}\frac{1}{n^\alpha(\ln n)^\beta} $$ converges if and only if $\alpha>1$, or $\alpha=1$ and $\beta>1$.

In your case $\alpha=0$ so it always diverges.

Proof in the case $\alpha =0$ and $\beta>0$ (your case).:

Since the general term is nonnegative and nonincreasing, the series is of the same nature as the the integral $$ \int_2^{+\infty}\frac{1}{(\ln x)^\beta}dx. $$ Now observe that $$ \lim_{x\rightarrow+\infty}\frac{\sqrt{x}}{(\ln x)^\beta}=+\infty. $$ So eventually, for $x\geq M\geq $, say, $$ \frac{1}{(\ln x)^\beta}\geq \frac{1}{\sqrt{x}} $$ hence $$ \int_2^{+\infty}\frac{1}{(\ln x)^\beta}dx\geq \int_M^{+\infty}\frac{1}{(\ln x)^\beta}dx \geq \int_M^{+\infty}\frac{1}{\sqrt{x}}dx=+\infty. $$

edited Feb 13 '13 at 14:18

answered Feb 13 '13 at 14:04

Julien

44,791

Is there not easier example? Without logarithm? – mathvisitor Feb 13 '13 at 14:05
That's a very natural example you found. I can't think of an easier example right now, and I doubt you can find one. Simpler, for example, would be Riemann $p$-series, but they don't help here. Note that it depends alos on what you call easier. Bertrand series are not that difficlt, after all. – Julien Feb 13 '13 at 14:20

Sequences not in $l^p$

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