The full question wouldn't fit in the title. Here it is:
Let ($a_n$)$_n$$_\in$$_\Bbb{N}$ and ($b_n$)$_n$$_\in$$_\Bbb{N}$ be two sequences and suppose that the set {n$\in$$\Bbb{N}$: $a_n\neq b_n$} is finite ($a_n$ and $b_n$ differ for finitely many values of n). Prove that either both sequences converge to the same limit or both diverge.
Our definition for convergence is: Given a real number $L$, we say that $(X_n)$ converges to L if for every $\epsilon$>0, there exists N∈$\Bbb{N}$ such that for all n∈N satisfying n>N, we have |$X_n-L$|<$\epsilon$. I'm just not sure if I'm supposed to apply that definition to this problem and if so, then how?
Best Answer
If the set $\{n\in \mathbb N: a_n \neq b_n\}$ is finite, there is an $N_0$ such that $a_n=b_n \ \forall n>N_0$. Now, suppose that $\lim_{n} a_n=L$ exists and it's a real number. Then,
$$\forall \varepsilon>0 \quad \exists N\in \mathbb N: n>N \Rightarrow |a_n-L| <\varepsilon$$ and now we'd like to prove that $\lim_{n} b_n=L$, that is, $$\forall \varepsilon>0 \quad \exists M\in \mathbb N: n>M \Rightarrow |b_n-L| <\varepsilon$$ but if we choose $M=\max\{N, N_0\}$, then $$n>M \Rightarrow |b_n-L|=|a_n-L|<\varepsilon$$