# If $\lim_{n\to \infty} (a_n - b_n) = 0$, with $\{a_n\}$ converging, then $\{b_n\}$ converges to the same limit.

Here is a question I am working on:

Let $\{a_n\}$ be a sequence with limit $L$. Suppose that $\{b_n\}$ is a sequence such that for some positive integer $N$ we have $b_n$ = $a_n$ for every $n\ge N$. Prove that $\lim_{n\to \infty} b_n = L$.

Here is my solution, I think it seems too simple, but I would appreciate some feedback or some tips on a better direction:

We have that $b_n$ = $a_n$ for every $n\ge N$. Simply take the limit of both sides to yield:

$\lim_{n\to \infty}b_n = \lim_{n\to \infty}a_n = L.$

Hence $\lim_{n\to \infty} b_n = L$, as required.

• "Simply take the limit of both sides" I don't think this is the purpose for this exercise. – Jack Nov 28 '17 at 22:57
• You are just asserting that it is true because it is true. – José Carlos Santos Nov 28 '17 at 22:57
• if $a_n-b_n=0$ then $a_n=b_n$ so they are the same sequence. I think the title is wrong, you want surely to write that $\lim_{n\to\infty}(a_n-b_n)=0$ instead – Masacroso Nov 28 '17 at 23:16
• Thanks! I'll change the title. Sorry about that – elysecol Nov 28 '17 at 23:17
• I think you now need to change the question to match the title. This can be done as any limit problem assuming $L <$ infinity. Let $\epsilon$ be an arbitrary (small) real number. Using the two limits you know to show that for $n$ large enough ... – Stephen Meskin Nov 28 '17 at 23:25

By the definition of limit you have that for every $\varepsilon>0$, there exists $k\in \mathbb{N}$ such that $|a_n-L|<\varepsilon$, whenever $n\geq k$. By hypothesis, there is $N \in \mathbb{N}$ such that $a_m=b_m$ for $m\geq N$. Taking $K:=\max\{N,k\}$ yields $$|a_n-L|=|b_n-L|<\varepsilon,\: \forall n\geq K;$$ this finishes the proof.