$x=\left (1,\frac{1}{\sqrt{2}\ln(2)},\frac{1}{\sqrt{3}\ln(3)},\dots,\frac{1}{\sqrt{n}\ln(n)},\dots\right)\in l_p$?

Question

Show that: $$x=\left(1,\frac{1}{\sqrt{2}\ln(2)},\frac{1}{\sqrt{3}\ln(3)},\dots,\frac{1}{\sqrt{n}\ln(n)},\dots\right)\in c_0$$

belongs to the space $l_p$ for $p\geqslant 2 $

I tried to write down the sum like this $\sum_\limits{n=1}^{\infty}\frac{1}{\sqrt{n}\ln(n)}$, the problem is that when $n=1$, the series diverges to infinity. I have no clue how to write these series so that it does not diverge. Even if I used $l_2$ norm it would still diverge at $1$.

Question:

How should I solve the exercise?

Thanks in advance!

Answer 1

Hint. Note that for $p\geq 2$ $$\sum_{n=3}^{\infty}\frac{1}{n^{p/2}\ln^p(n)}\leq \sum_{n=3}^{\infty} \frac{1}{n\ln^2(n)}\leq \sum_{n=3}^{\infty} \int_{x=n-1}^{n}\frac{1}{x\ln^2(x)}\,dx = \int_{x=2}^{\infty}\frac{1}{x\ln^2(x)}\,dx.$$

Answer 2

There exists $k\in \Bbb N$ such that $\forall n\geq k\;(\;0<\sqrt n\;\log n <\sqrt {n+1}\;\log (n+1)\;)$ so apply the Cauchy Condensation Test . This test also shows that $x\not \in l_p$ for $1\leq p<2.$