Showing that a series in $l_{\infty}$ converges weakly, given a boundedness condition. I'm trying to understand the following:

Let $\{x_k\}_{k=1}^{\infty}$ be a sequence of elements in $l_{\infty}$ such that for some constant $K$, $$\|\sum_{1}^n \lambda_k x_k \|\leq K\sup_k |\lambda_k| \quad \textrm{for all} \ \{\lambda_k\}_{k=1}^n\subset \mathbb{R}, \quad n=1,2,\dots$$ Then for every bounded sequence $\{\lambda\}_{k=1}^{\infty} \subset \mathbb{R}$ the series $\sum_1^{\infty} \lambda_k x_k$ converges weakly to an element of norm $\leq K \sup_k |\lambda_k |$ in $l_{\infty}$.

From the stated condition we can say that the $\sup$ of the inequality's LHS is finite. Does that imply the convergence of the series? (Which would mean strong convergence, implying weak convergence) If not, how can I convince myself of the weak convergence?

Let $M=\sup_{k\ge 1} |\lambda_k|$. By assumption, the partial sums $s_n=\sum_{k=1}^n \lambda_k x_k$ satisfy  $\|s_n\|_\infty\le MK$ for all $n$. For bounded sequences, weak* convergence in $\ell_\infty$  is precisely the coordinate-wise convergence. Thus, we only need to check that for each fixed index $j$ the numeric series $\sum_{k=1}^\infty \lambda_k x_k(j)$ converges. 
Fix $j$. For each $k$, choose $\mu_k \in\{1,-1\} $ so that  $\mu_k x_k(j) =  |x_k(j)|$.
By assumption, the partial sums of the series $\sum_{k=1}^\infty \mu_k x_k(j) $ are bounded by $K$, which implies 
$\sum_{k=1}^\infty |x_k(j)|$ converges to a number at most $K$. Thus, 
$\sum_{k=1}^\infty \lambda_k x_k(j)$ converges and
$$\left|\sum_{k=1}^\infty \lambda_k x_k(j )\right| \le M\sum_{k=1}^{\infty} |x_k(j)| \le MK$$
