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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!

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    $\begingroup$ George.Suggestion: $|b_n-L| \le |b_n-a_n| +|a_n-L|$.Given $\epsilon/2$ there is a $n_0$ for $|a_n-L|$. $|b_n-a_n|<r2^{-n}$.Since $r2^{-n}$ converges to zero there is a $n_1$ s.t.for $n \ge n_1$: $r2^{-n}<\epsilon/2$.Take $\max (n_0,n_1)$.ok? $\endgroup$ Feb 17, 2020 at 22:31
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    $\begingroup$ George. No worry. I would start with one!! epsilon, not introduce $\epsilon_1$, and $\epsilon_2$. For this $\epsilon/2$ you get $n_0,$ and $n_1$ for the two parts. Max will ensure you can use each individual inequality. And the you have $\epsilon/2+\epsilon/2=\epsilon$, and you are done.ok? $\endgroup$ Feb 17, 2020 at 22:52
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    $\begingroup$ A general advice in such cases, if you like, is to start from what you w a n t to prove (here from $(b_n)$). This will give you o n e $\varepsilon$ and you will carry this positive number in order to use it in the definitions of the limits of $a_n\to a$ and $b_n-a_n\to 0$. Have in mind that whenever you know that f o r a l l $\varepsilon>0$ something happens, then it is very common(!) to use a specific $\varepsilon$ that will lead you to the desired result (here $\varepsilon/2$ in both cases).. $\endgroup$ Feb 17, 2020 at 22:59
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    $\begingroup$ George. Just write it down, the whole thing, with one epsilon (actually for aesthetical reasons one chooses epsilon/2) and see that the proof is simpler than your 2 epsilon proof. Keep it simple. Cheers. $\endgroup$ Feb 17, 2020 at 23:05
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    $\begingroup$ @GeorgeCooper The problem is in the line $\epsilon_1<\epsilon_2-2^{-n}r$. You can't, in general, choose $\epsilon_1$ in such a manner. Suppose that instead of $2^{-n}$, you had $2^n$ in the problem. Nothing changes in your proof, and you reach the same conclusion which would be wrong $\endgroup$
    – bjorn93
    Feb 17, 2020 at 23:28

2 Answers 2

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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

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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.

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