Show $\lim\limits_{n \to \infty} x_n=a$

Question

Given $n,k \in \mathbb{N}$, $t_{n,k} \geq 0$, $\sum\limits_{k=1}^n t_{n,k}=1$, $\lim\limits_{n \to \infty}t_{n,k}=0$. $\lim\limits_{n \to \infty}a_n=a$ and let $x_n := \sum\limits_{k=1}^n t_{n,k}a_k$. Show $\lim\limits_{n \to \infty} x_n=a$.

I think I can see why intuitively. For large $n$, in $\sum\limits_{k=1}^n t_{n,k}a_k$, we are summing big portion of term of the form $\epsilon_j (a\pm\epsilon_i)$, so it is approximately $(1-\epsilon)a+\epsilon*First\_few\_term\_of\_a_n \approx a$.

Is there clever ways to prove it?

Answer 1

Note that the sequence $x_n$ forgets the 'past' values of $a_k$ at the infinity by observing $$ \lim_{n\to\infty}\sum_{k=1}^M t_{n,k}a_k = 0 $$ for all fixed $M>0$. This enables us to assume without loss of generality, $|a_n-a|<\epsilon$ for all $n$, giving the desired result. So the formal proof goes like this. Choose $M>0$ such that $$ |a_n-a|<\epsilon,\quad \forall n>M. $$ Then we have $$ |x_n -a| \leq \sum_{k=1}^M t_{n,k}|a_k-a| + \epsilon\sum_{k=M+1}^n t_{n,k}<\sum_{k=1}^M t_{n,k}|a_k-a|+ \epsilon,\quad\forall n>M. $$ Take $n\to \infty$ to get $$ \limsup_{n\to\infty} |x_n -a|\leq \epsilon, $$ for arbitrary $\epsilon>0$. Conclude from this that $\lim_{n\to\infty} x_n =a$.