Let $\{a_k\}$ be a sequence of non negative real numbers such that $\sum_{k=1}^{\infty} a_k =1$. Let $b_{nk}$ is a double array sequence of real numbers where $n \in \mathbb{N}$.It is given that $0 \le b_{nk} \le M$ uniformly. Suppose $\lim_{n \rightarrow \infty} b_{nk}=c_k $ for all $k$.Prove that $\lim_{n \rightarrow \infty}\sum_{k=1}^{\infty}a_k b_{nk}=\sum_{k=1}^{\infty}a_kc_k$
Now, I approached this by looking at the difference $\sum_{k=1}^{\infty} a_k b_{nk} - a_k c_k=\sum_{k=1}^{\infty} a_k (b_{nk}-c_k)$.
As we know that $|b_{nk}-c_k|< \epsilon$ for $n > N_{k}$. I take the supremum of all such $N_k$'s to be $N$. So, if now $n>N$, the difference must go to $0$ right? But how can I guarantee the existence of the supremum? Also, how can I use the boundedness of $b_{nk}$?
Hint: Given $\epsilon >0$ choose $N$ such that $\sum\limits_{k=N+1}^{\infty} a_k <\epsilon$. Note that $|\sum\limits_{k=N+1}^{\infty} [a_kb_{nk} -a_kc_k]| \leq 2M\epsilon$ since the hypothesis implies that $|c_k| \leq M$ for all $k$ which gives $|b_{nk}-c_k| \leq 2M$. Next note that $|\sum\limits_{k=1}^{N} [a_kb_{nk} -a_kc_k]| \to 0$ as $n \to \infty$. I will let you finish the proof.