Sum of product of bounded and convergent sequences I have four sequences as $f_{i,k}$ where $k\in \{1,2,3,4\}$ and each of them is convergent, i.e. $\lim_{i \to \infty} f_{i,k} = f_{k}$ for $k\in \{1,2,3,4\}$ and $f_{m} \neq f_{n}$ where $m,n \in \{1,2,3,4\}, m \neq n$. I have four other sequences as $\lambda_{i,k} \in [0,1]$ and $\sum_{k=1}^{4} \lambda_{i,k} = 1$ for each i. I know that $\lim_{i \to \infty} \sum_{k=1}^{4} \lambda_{i,k}f_{i,k}$ is convergent to something. What can we say about $\lambda_{i,k}$ sequences? Can we say $\lim_{i \to \infty} \lambda_{i,k} = \lambda_{k}^{*}$ where $\sum_{k=1}^{4} \lambda_{k}^{*} = 1$ and $\lambda_{k}^{*} \in [0,1]$?
Thanks!
 A: I consider the $f_{i,k}=\vec{f}_i$ and $\lambda_{i,k}=\vec{\lambda}_i$ to be sequences in $\Bbb R^K$, and the limit point $f$ is a point in $\Bbb R^K$. In your case $K=4$.
It would be ideal if the $\vec{\lambda}_i$ converged to a limit vector $\vec{\mu} = (\mu_1,\mu_2,\dots,\mu_K)$ which satisfied the following system:
$$ \begin{cases}
\sum_{k\geq1} \mu_{k} & = & 1 \\
\sum_{k\geq1} \mu_{k} f_k & = & L
\tag{1} \end{cases} $$
where $L=\lim_{i\rightarrow\infty} \sum_{k\geq1} \lambda_{i,k} f_{i,k}$. This is sort of the case. The system has unique solutions for $K\leq 2$, and infinitely many solutions for $K\geq 3$. 
For $K=1$ and $2$, it can be shown that $\vec{\lambda}_i$ must converge to the unique solution of the system $(1)$. For $K\geq 3$, however, convergence is not guaranteed. Let $\vec{\mu}^{(1)}$ and $\vec{\mu}^{(2)}$ be two solutions to $(1)$. Then $\lambda_{i,k}$ can be defined as follows:
$$ \vec{\lambda}_{i} = \begin{cases}
\vec{\mu}^{(1)} & \text{if } i \text{ is odd} \\
\vec{\mu}^{(2)} & \text{if } i \text{ is even} \\
\end{cases} $$
This sequence satisfies the system $(1)$ but it does not converge, proving that convergence is not guaranteed by $(1)$. However, you can guarantee that the sequence $\vec{\lambda}_i$ is getting arbitrarily close to the set $S$ of all solutions $\vec{\mu}$ of the system $(1)$, ie.
$$ \lim_{i\rightarrow\infty}d(\vec{\lambda}_i,S) = \lim_{i\rightarrow\infty}\left(\inf\{ d(\vec{\lambda}_i,\vec{\mu}) : \vec{\mu}\in S\}\right) = 0$$
