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

Solution 1:

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

Solution 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 $0$.