If $\lim_{n\to\infty}(a_n)=\alpha\in\mathbb R$ and $|b_n-a_n|\leq 2^{-n}r$. Show that $\lim_{n\to\infty}(b_n)= \alpha$. Question:
Let $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ be two real sequences with $\lim\limits_{n\to\infty}(a_n)=\alpha\in\mathbb R$ and $|b_n-a_n|\leq 2^{-n}r$ where $r\gt 0$. Show that $(b_n)_{n\in\mathbb N}$ converges to $\alpha$.
My answer:
If $\lim\limits_{n\to\infty}(a_n)=\alpha \Leftrightarrow \forall \epsilon_1\gt0 \space$ $\exists N_1\in\mathbb N$ : $n>N_1 \space\Rightarrow\space |a_n-\alpha|\lt\epsilon_1.$
We know $|b_n-a_n|\le2^{-n}r,\space r\gt0.$
$\Rightarrow\space-2^{-n}r\le b_n-a_n \le2^{-n}r\space\space$ & $\space\space\alpha-\epsilon_1\lt a_n\lt\alpha+\epsilon_1$
$\Rightarrow \alpha-\epsilon_1-2^{-n}r\lt b_n\lt\alpha+\epsilon_1+2^{-n}r$
$\Rightarrow -\epsilon_1-2^{-n}r\lt b_n-\alpha\lt\epsilon_1+2^{-n}r$
$\Rightarrow |b_n-\alpha|\lt\epsilon_1+2^{-n}r$
We want to show: $\forall \epsilon_2\space\exists N_2\in\mathbb N:\space n\gt N_2 \space \Rightarrow \space |b_n-\alpha|\lt \epsilon_2$
Choosing $\epsilon_1 \lt \epsilon_2-2^{-n}r,\space N_2=N_1 \space \Rightarrow \space \alpha-\epsilon_1\lt a_n\lt\alpha+\epsilon_1\space; \space$ we also know that $|b_n-a_n|\le2^{-n}r$
Using this information, we get:
$|b_n-\alpha|\lt\epsilon_1+2^{-n}r \lt \epsilon_2$
Hence $\lim\limits_{n\to\infty}(b_n)=\alpha$
I'm unsure if I am allowed to choose $\epsilon_1$ and $N_2$ so freely. I was just wondering if someone, who is more familiar with this topic, could check my logic (I have just started Analysis I) even if I have completely messed it up!
 A: Note that $\epsilon_2-2^{-n}r$ could be negative (where your proof breaks down), unless you control the term $2^{-n}r$ suitably.
A possible proof using only one $\epsilon$ goes as follows:
Given the assumptions, one proves that $\lim_{n\rightarrow \infty}b_n=\alpha$.
Proof: $\forall \epsilon>0,\exists N_1$ such that $$|a_n-\alpha|<\frac {\epsilon}2,\forall n>N_1.$$ Furthermore $\exists N_2$ such that $$|b_n-a_n|\leq 2^{-n}r<\frac {\epsilon}2,\forall n>N_2.$$ Now take $N=\max(N_1,N_2)$. Then $$|b
_n-\alpha|=|(b_n-a_n)+(a_n-\alpha)|\leq|b_n-a_n|+|a_n-\alpha|<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon,\forall n>N.$$ QED
A: The key here is that $r/2^n\to 0$ as $n\to\infty $. If you are allowed to use this fact then the proof is a bit shorter. Otherwise let's note that $2^n>1+n$ for $n>1$ so that $r/2^n<r/(n+1)$ and hence for a given $\epsilon>0$ we have a positive integer $N_1=\lfloor 2r/\epsilon \rfloor$ such that $0<r/2^n<\epsilon/2$ whenever $n>N_1$.
By given assumption there is another positive integer $N_2$ such that $|a_n-\alpha|<\epsilon/2$ whenever $n>N_2$. Therefore if $n>N=\max(N_1,N_2)$ then we have $$|b_n-\alpha|\leq |b_n-a_n|+|a_n-\alpha|$$ which is less than $r/2^n+\epsilon /2$ and therefore less than $\epsilon$. It follows that $b_n\to\alpha$ as $n\to\infty $.

You should now have understood that your result works when $r/2^n$ is replaced by any sequence tending to zero.
Often such epsilon based proofs are not needed. Instead one is supposed to use limit theorems (which are already proved using epsilon stuff). Here you can directly use Squeeze theorem along with known information that $r/2^n\to 0$. Just note that inequality in question can be written as $$a_n-r/2^n\leq b_n\leq a_n+r/2^n$$ and leftmost and rightmost expressions of the above inequality tend to $\alpha$ so that by Squeeze theorem the middle term $b_n$ also does the same. 
