Recall that a bounded subset $A$ of a Banach space $X$ is said to be weakly precompact if every sequence in $A$ admits a weakly Cauchy subsequence. Rosenthal's $l_{1}$-theorem states that a bounded subset $A$ is weakly precompact if and only if it contains no $l_{1}$-sequence. For a bounded subset $A$ of $X$, we let $$\operatorname{wpc}_{X}(A)=\inf\{\epsilon>0:A\subseteq K_{\epsilon}+\epsilon B_{X}\},$$ where $K_{\epsilon}$ is weakly precompact.
It is easy to see that $\operatorname{wpc}_{X}(A)=0$ if and only if $A$ is weakly precompact. Clearly, every weakly Cauchy sequence is weakly precompact. Next I want to know how to quantify this simple fact.
Let $(x_{n})_{n}$ be a bounded sequence of $X$. We let $$\delta((x_{n})_{n})=\sup_{x^{*}\in B_{X^{*}}}\inf_{n}\sup_{k,l\geq n}|\langle x^{*},x_{k}-x_{l}\rangle|.$$ Clearly, $\delta((x_{n})_{n})=0$ if and only if $(x_{n})_{n}$ is weakly Cauchy. It is easy to see that $\delta((x_{n})_{n})$ is the diameter of the set of all $weak^{*}$-cluster points of $(x_{n})_{n}$ in $X^{**}$.
Question. $\operatorname{wpc}_{X}((x_{n})_{n})\leq \delta((x_{n})_{n})$ ?
Thank you !