Recall that a series $(\sum_{k=1}^{n} x_{k})_{n\in\mathbb{N}}$ in a Banach space $X$ is called weak unconditionally Cauchy if for every $x^{*}\in X^{*}$ the series $(\sum_{k=1}^{n} x^{*}(x_{k}))_{n\in\mathbb{N}}$ in $\mathbb{K}$ is unconditionally Cauchy. Recall that a series in $\mathbb{K}$ is unconditionally Cauchy if and only if it is absolutely convergent.
Claim. Let $X$ be a Banach space and $(x_{n}^{*})_{n\in\mathbb{N}}$ a sequence in $X^{*}$. The following are equivalent.
$({\rm i})$ The series $(\sum_{k=1}^{n} x_{k}^{*} )_{n\in\mathbb{N}}$ in $X^{*}$ is weak unconditionally Cauchy.
$({\rm ii})$ There is a bounded linear operator $T\colon X\to \ell_{1}$ such that $Tx = (x_{n}^{*}(x))_{n\in\mathbb{N}}$ for all $x\in X$.
Proof. $({\rm i}) \Rightarrow ({\rm ii})$. Suppose the series $(\sum_{k=1}^{n} x_{k}^{*} )_{n\in\mathbb{N}}$ in $X^{*}$ is weak unconditionally Cauchy. Let $x\in X$. Since the map $f\mapsto f(x)$ defined on $X^{*}$ is a continuous linear functional on $X^{*}$, the series $(\sum_{k=1}^{n} x_{k}^{*}(x))_{n\in\mathbb{N}}$ is absolutely convergent. Hence $(x_{n}^{*}(x))_{n\in\mathbb{N}}$ belongs to $\ell_{1}$. Define $T\colon X\to \ell_{1}$ by
\begin{equation}
Tx := (x_{n}^{*}(x))_{n\in\mathbb{N}} .
\end{equation}
Observe that $T$ is a well-defined linear operator. A standard argument shows that $T$ has a closed graph. We include the argument for completeness. Let $(x_{k})_{k\in\mathbb{N}}$ be a sequence in $X$, $x\in X$ and $(\alpha_{n})_{n\in\mathbb{N}} \in \ell_{1}$ such that $(x_{k})_{k\in\mathbb{N}}$ converges to $x$ in $X$ and $(Tx_{k})_{k\in\mathbb{N}}$ converges to $(\alpha_{n})_{n\in\mathbb{N}}$ in $\ell_{1}$. Let $n\in\mathbb{N}$. Since the sequence $(x_{k})_{k\in\mathbb{N}}$ converges to $x$ in $X$ and $x_{n}^{*}\in X^{*}$ we have
\begin{equation}
\lim_{k\to\infty} x_{n}^{*}(x_{k}) = x_{n}^{*}(x) . \tag{1}
\end{equation}
Now define $e_{n}^{*}\colon \ell_{1}\to\mathbb{K}$ by $e_{n}^{*}((\beta_{k})_{k\in\mathbb{N}} ) := \beta_{n}$. Then $e_{n}^{*}\in \ell_{1}^{*}$. As $(Tx_{k})_{k\in\mathbb{N}}$ converges to $(\alpha_{j})_{j\in\mathbb{N}}$ in $\ell_{1}$ and $e_{n}^{*}\in \ell_{1}^{*}$, we have
\begin{equation}
\lim_{k\to\infty} x_{n}^{*}(x_{k}) = \lim_{k\to\infty} e_{n}^{*}(Tx_{k}) = e_{n}^{*}((\alpha_{j})_{j\in\mathbb{N}} ) = \alpha_{n} . \tag{2}
\end{equation}
By combining $(1)$ and $(2)$ we obtain
\begin{equation}
x_{n}^{*}(x) = \alpha_{n} . \tag{3}
\end{equation}
As $n$ was arbitrary we see from $(3)$ that $Tx = (\alpha_{n})_{n\in\mathbb{N}}$. Hence $T$ has a closed graph. By the closed graph theorem we see that $T$ is bounded. Hence $T$ has the desired properties.
$({\rm ii}) \Rightarrow ({\rm i})$. Assume there is a bounded linear operator $T\colon X\to \ell_{1}$ such that $Tx = (x_{n}^{*}(x))_{n\in\mathbb{N}}$ for all $x\in X$. For each $n\in\mathbb{N}$ define $e_{n}^{*}\colon \ell_{1}\to\mathbb{K}$ by $e_{n}^{*}((\alpha_{j})_{j\in\mathbb{N}}) := \alpha_{n}$. Let $\Phi \colon \ell_{\infty} \to \ell_{1}^{*}$ be the usual isomorphism given by
\begin{equation}
(\Phi (\beta_{j})_{j\in\mathbb{N}}))(\alpha_{i})_{i\in\mathbb{N}} = \sum_{k\in\mathbb{N}} \alpha_{k} \beta_{k} .
\end{equation}
Define $S := (T^{*}\circ \Phi)\vert_{c_{0}}$. Observe that $S$ is a bounded linear operator from $c_{0}$ into $X^{*}$. Let $n\in\mathbb{N}$ and $x\in X$. We have
\begin{equation}
(Se_{n})(x) = (T^{*}(\Phi e_{n}))(x) = (\Phi e_{n})(Tx) = e_{n}^{*}(Tx) = x_{n}^{*}(x) .
\end{equation}
Hence we have $Se_{n} = x_{n}^{*}$ for each $n\in\mathbb{N}$. Let $x^{**}\in X^{**}$. Then $S^{*}x^{**}\in c_{0}^{*}$. Since the series $(\sum_{k=1}^{n} e_{k})_{N\in\mathbb{N}}$ is weak unconditionally Cauchy in $c_{0}$, the series $(\sum_{k=1}^{n} (S^{*} x^{**})e_{k})_{N\in\mathbb{N}}$ is absolutely convergent. Hence for each $N\in\mathbb{N}$ we have
\begin{equation}
\sum_{k=1}^{N} |x^{**}(x_{k}^{*})| = \sum_{k=1}^{N} |x^{**}(Se_{k})| = \sum_{k=1}^{N} |(S^{*}x^{**})e_{k}| \leq \sum_{k=1}^{\infty} |(S^{*}x^{**})e_{k}| < \infty . \tag{4}
\end{equation}
It follows from $(4)$ that the series $(\sum_{k=1}^{n} x^{**}(x_{k}^{*}))_{n\in\mathbb{N}}$ is unconditionally Cauchy. Hence $(\sum_{k=1}^{n} x_{k}^{*})_{n\in\mathbb{N}}$ is weak unconditionally Cauchy.
