Show $a_n, b_n$ converge to the same limit

Question

Let $a_n$, $b_n$ be two series, such that for every $\epsilon>0$ exists an natural N such that for every $n,m>N$ $|a_n-b_m|<\epsilon$ . Prove that both series converge and to the same limit.

I've been trying to show that they are both Cauchy sequences with the triangle inequality, but it doesn't seem to work. I think I know how to show the converge to the same limit but that only works assuming they both converge which I haven't managed to do. Any help would be appreciated.

Answer 1

Let $\epsilon > 0$. There is a $N \in \mathbb N$, such that for all $n, m \geq N$ we have $$ |a_n - b_m| < \frac{\epsilon}{2} \; . $$ Using the triangle inequality, we get that for all $n, m \geq N$ we have $$ |a_n - a_m| \leq |a_n - b_N| + |b_N - a_m| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \; , $$ so $(a_n)_n$ is Cauchy. We can use the same argument to show that $(b_n)_n$ is Cauchy. Now we know that these two sequences converge, so there is a limit $a$ for the sequence $(a_n)_n$ and a limit $b$ for the sequence $(b_n)_n$. We need to show, that $a = b$. Let's do that by using a contradiction argument. So assume that $a \neq b$, and let $\epsilon := |a-b| > 0$. There exists a $N \in \mathbb N$, such that for each $n, m \geq N$, the following three conditions hold: $$ \begin{align*} |a - a_n| &< \frac{\epsilon}{3} \; ,\\ |b - b_m| &< \frac{\epsilon}{3} \; ,\\ |a_n - b_m| &< \frac{\epsilon}{3} \; . \end{align*} $$ By using the triangle inequality two times, we get $$ \epsilon = |a - b| \leq |a-a_N| + |a_N - b_N| + |b_N - b| < \epsilon \; , $$ which is a contradiction, so we deduce that $a = b$.

Answer 2

For $\epsilon >0$, and $n, m \ge N$, take $n \ge m$ then

$\mid b_n - b_m \mid \le \mid a_n - b_n\mid + \mid a_n - b_m\mid <2\epsilon$. So, $\{b_n\}$ is cauchy. Similarly for $\{a_n\}$.

Answer 3

I think you are looking to show that if $b_n$ is a subsequence of $a_n$, they both converge to the same limit. If so, the proof is as follows.

let $lim_n a_n = l$ and let $\epsilon >0$ be given.

There exists some index $N \in \mathbb{N}$ such that $|{a_n-l}| < \epsilon$ for all $n\geq N$. This index $N$ will also work for the subsequence $b_n$, since if $n \geq N$ then $b_n = a_m$ for some $m \geq n \geq N$ so that $|b_n - l| = |a_m - l| < \epsilon$

Answer 4

In addition to Mr.X: Moreover, let $b$ be the limit of the sequence $(b_n)$ and we want to show $a_n \to b$. For $\varepsilon>0$ choose $N\in \mathbb{N}$ such that $|b_m-b|<\varepsilon/2$ for $m>N$ and $|a_n-b_m|<\varepsilon/2$ for all $n,m>N$. This shows the desired convergence. For $(b_n)$ the same trick applies.