Is there Banach space $X$ and non Cauchy sequence $(x_n)\subset X$ which satisfy $$ \forall_{x^*\in X^*} \exists_{L>0}|x^*(x_n)|<\frac{L}{\sqrt{n}}? $$
I already tried finding such sequence in $L^P$ and $l^p$ spaces and found sequence which satisfies this inequality for almost every $x^*\in X^*$. That is $X=L^2(0,1)$ and $x_n=\sqrt{n}\chi_{(0,\frac{1}{n})}$. By Holder inequality for $f\in L^2(0,1)\cap L^{\infty}(0,1)$ we got $$ |f(x_n)|\leq||f||_{\infty}\cdot||x_n||_1=\frac{||f||_{\infty}}{\sqrt{n}}<\frac{||f||_{\infty}+1}{\sqrt{n}}. $$ Is there a way to show the inequality holds for all $f\in L^2(0,1)$? Or should I focus on another space?
EDIT: As Joey pointed in comment this is not a good example. Here is my another approach.
Let $(x^{(n)})_{n=1}^{\infty}\subset \ell^2$ such that $$ x^{(n)}_k=\begin{cases} \frac{1}{2^{m/2}} & \text{if }k=mn\text{ for some } m\in\mathbb{N} \\ 0 & \text{otherwise} \end{cases}. $$ Then $||x^{(n)}||_2=1$ for each $n\in\mathbb{N}$ and for $f\in\ell^2$ we get $$ |f(x^{(n)})|= \Big|\sum_{k=1}^{\infty}f_{kn}\frac{1}{2^{k/2}}\Big|. $$ Now using Holder inequality we obtain $$ \Big|\sum_{k=1}^{\infty}f_{kn}\frac{1}{2^{k/2}}\Big|\leq \sum_{k=1}^{\infty}\Big|f_{kn}\frac{1}{2^{k/2}}\Big|\leq \Big(\sum_{k=1}^{\infty}|f_{kn}|^2\Big)^{\frac12} \Big(\sum_{k=1}^{\infty}\Big|\frac{1}{2^{k/2}}\Big|^2\Big)^{\frac12}. $$ Here the second sum is equal 1. Now lets focus on first sum without square root. We can estimate it using $\sup$ $$ \sum_{k=1}^{\infty}|f_{kn}|^2\leq \sum_{k=1}^{\infty}\sup_{m\in ((k-1)n, kn]}|f_m|^2. $$ And i can't get any further. My intuition is that we sum $\sup$ on intervals of length $n$, so the sum should be less than $C||f||_2^2/n$.
As user58955 wrote in the comments, if we have a Banach space $X$ and a sequence $(x_n) \subseteq X$ satisfying
$$\forall_{x^*\in X^*} \exists_{L>0}|x^*(\sqrt{n}x_n)|<L,$$
then by the uniform boundedness principle the sequence $(\sqrt{n} x_n)$ is bounded in $X$, i.e. $\exists_{L>0} ||\sqrt{n} x_n|| \leq L$ for all $n \in \mathbb{N}$. But then $||x_n|| \leq \frac{L}{\sqrt{n}} \rightarrow 0$. Hence, $x_n \rightarrow 0$ in norm and is a Cauchy sequence.