Let $E$ be a $\mathbb R$-Banach space and $(x_n)_{n\in\mathbb N}\subseteq E$. Assume $\left(\sum_{k=1}^Kx_{n_k}\right)_{K\in\mathbb N}$ is weakly convergent for all increasing $(n_k)_{k\in\mathbb N}\subseteq\mathbb N$. Note that $(\varphi(x_n))_{n\in\mathbb N}$ is absolutely summable for all $\varphi\in X'$ and hence $$L\varphi:=(\varphi(x_n))_{n\in\mathbb N}\;\;\;\text{for }\varphi\in X'$$ is a linear operator from $X'$ to $\ell^1$?
How can we show that $L$ is closed?
Let $(\varphi_m)_{m\in\mathbb N}\subseteq X'$ with $$\left\|\varphi-\varphi_m\right\|_{X'}\xrightarrow{m\to\infty}0\tag1$$ for some $\varphi\in X'$ and $$\left\|(y_n)_{n\in\mathbb N}-L\varphi_m\right\|_{\ell^1}\xrightarrow{m\to\infty}0\tag2$$ for some $(y_n)_{n\in\mathbb N}\in\ell^1$. We need to show that $L\varphi=(y_n)_{n\in\mathbb N}$, i.e. $$\varphi(x_n)=y_n\tag3\;\;\;\text{for all }n\in\mathbb N\;.$$ Using $$\left\|L\varphi-(y_n)_{n\in\mathbb N}\right\|_{\ell^1}\le\left\|L\varphi-L\varphi_m\right\|_{\ell^1}+\left\|(y_n)_{n\in\mathbb N}-L\varphi_m\right\|_{\ell^1}\;,\tag4$$ it seems like we need to show $$\sum_{n\in\mathbb N}|(\varphi-\varphi_m)(x_n)|=\left\|L\varphi-L\varphi_m\right\|_{\ell^1}\xrightarrow{m\to\infty}0\;.\tag5$$ Clearly, $$|(\varphi-\varphi_m)(x_n)|\le\left\|\varphi-\varphi_m\right\|_{X'}\left\|x_n\right\|_E\tag6\;,$$ but since $(x_n)_{n\in\mathbb N}$ is not known to be absolutely summable (and hence $\sum_{n\in\mathbb N}\left\|x_n\right\|_E$ might be $\infty$), I don't see how we can conclude.
Let $y,g_m,g:\mathbb{N}\to\mathbb{R}$ defined by $$g_m(n) = \varphi_m(x_n),\quad g(n)=\varphi_m(x_n),\quad y(n)=y_n$$ and see $\mathbb{N}$ as a measure space with the counting measure $\mu$. Note that $g_m,g,y\in L^1(\mathbb{N})$.
Now, the hypothesis $\varphi_m\to\varphi$ in $\|\cdot\|_*$ implies $g_m\to g$ a.e. and $L\varphi_m\to y$ in $l^1$ means $g_m\to y$ in $L^1(\mathbb{N})$. Then, there is a subsequence $g_{m_k}$ such that $g_{m_k}\to y$ a.e.. So, $y=g$.