Proving "Limits of Finite Sequences Implies Limit of Sum

[Mr Davis 97]

Homework Statement

For each $n\in\mathbb{N}$, let the finite sequence $\{b_{n,m}\}_{m=1}^n\subset(0,\infty)$ be given. Assume, for each $n\in\mathbb{N}$, that $b_{n,1}+b_{n,2}+\cdots+b_{n,n}=1$.

Show that $\lim_{n\to\infty}( b_{n,1}\cdot a_1+b_{n,2}\cdot a_2+\cdots+b_{n,n}\cdot a_n) = \lim_{n\to\infty}a_n$, for every convergent sequence $\{a_n\}_{n=1}^\infty\subset\mathbb{R}$ if and only if, for each $m\in\mathbb{N}$, $\lim_{n\to\infty}b_{n,m}=0$.

Homework Equations

The Attempt at a Solution

I really need at least one hint for this one. Which direction of the proof should I start with? Which one is easier?

My idea for the <---- direction is that since for each $m\in\mathbb{N}$, $\lim_{n\to\infty}b_{n,m}=0$, we see that $\lim_{n\to\infty}b_{n,1}\cdot a_1+\lim_{n\to\infty}b_{n,2}\cdot a_2+\cdots+\lim_{n\to\infty}b_{n,n}\cdot a_n) = 0 + 0 + \cdots + 0$... But I'm not sure where this gets me.

 

[Mr Davis 97]

Homework Statement

For each $n\in\mathbb{N}$, let the finite sequence $\{b_{n,m}\}_{m=1}^n\subset(0,\infty)$ be given. Assume, for each $n\in\mathbb{N}$, that $b_{n,1}+b_{n,2}+\cdots+b_{n,n}=1$.

Show that $\lim_{n\to\infty}( b_{n,1}\cdot a_1+b_{n,2}\cdot a_2+\cdots+b_{n,n}\cdot a_n) = \lim_{n\to\infty}a_n$, for every convergent sequence $\{a_n\}_{n=1}^\infty\subset\mathbb{R}$ if and only if, for each $m\in\mathbb{N}$, $\lim_{n\to\infty}b_{n,m}=0$.

Homework Equations

The Attempt at a Solution

I really need at least one hint for this one. Which direction of the proof should I start with? Which one is easier?

My idea for the <---- direction is that since for each $m\in\mathbb{N}$, $\lim_{n\to\infty}b_{n,m}=0$, we see that $\lim_{n\to\infty}b_{n,1}\cdot a_1+\lim_{n\to\infty}b_{n,2}\cdot a_2+\cdots+\lim_{n\to\infty}b_{n,n}\cdot a_n) = 0 + 0 + \cdots + 0$... But I'm not sure where this gets me.

Here is my stab at a solution for the <--- direction.

Let $(a_n)$ be an arbitrary convergent sequence that converges to $c$. We want to show that $b_{n,1}a_n + \cdots b_{n,n}a_n$ also converges to $c$. To this end, given $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that if $n \ge N$ we have $|a_n - c| \le \epsilon$. $|b_{n,1} a_1 + \cdots b_{n,n}a_n - c| = |b_{n,1} a_1 + \cdots + b_{n,n}a_n - c(b_{n,1}+ \cdots + b_{n,n})| = |b_{n,1}(a_1 - c) + \cdots b_{n,n}(a_n - c)|$... This is kind of where I get stuck. I'm not sure what to do next.

 

[mfb]

My idea for the <---- direction is that since for each $m\in\mathbb{N}$, $\lim_{n\to\infty}b_{n,m}=0$, we see that $\lim_{n\to\infty}b_{n,1}\cdot a_1+\lim_{n\to\infty}b_{n,2}\cdot a_2+\cdots+\lim_{n\to\infty}b_{n,n}\cdot a_n) = 0 + 0 + \cdots + 0$... But I'm not sure where this gets me.

You cannot split a limit like that.

The other direction is easier if you make a proof by contradiction. Assume there is an m such that $\lim_{n\to\infty}b_{n,m} \neq 0$ and show that the original statement is no longer true. That should also give an idea how the overall structure works.