MHB Proof of $\lim_{{n}\to{\infty}}a_n=\lim_{{n}\to{\infty}}b_n$ with $\epsilon-N$

[Dethrone]

Let ${a_n}$ and ${b_n}$ be two convergent sequences such that $\lim_{{n}\to{\infty}}|a_n-b_n|=0$. Give an $\epsilon-N$ proof to show that $\lim_{{n}\to{\infty}}a_n=\lim_{{n}\to{\infty}}b_n$.

Right now, I've just said without loss of generality, assume that $a_n>b_n$ for all $n>N$, then since $\lim_{{n}\to{\infty}}|a_n-b_n|=0$, we have $\lim_{{n}\to{\infty}}a_n-\lim_{{n}\to{\infty}}b_n=0$ and the results follow.
But I haven't used "$\epsilon$" at all...I'm thinking I need to prove that $\forall \epsilon >0$, $\exists N$ s.t whenever $n>N, |a_n-\lim_{{n}\to{\infty}}b_n|=\epsilon$.

 

[Sudharaka]

Let ${a_n}$ and ${b_n}$ be two convergent sequences such that $\lim_{{n}\to{\infty}}|a_n-b_n|=0$. Give an $\epsilon-N$ proof to show that $\lim_{{n}\to{\infty}}a_n=\lim_{{n}\to{\infty}}b_n$.

Right now, I've just said without loss of generality, assume that $a_n>b_n$ for all $n>N$, then since $\lim_{{n}\to{\infty}}|a_n-b_n|=0$, we have $\lim_{{n}\to{\infty}}a_n-\lim_{{n}\to{\infty}}b_n=0$ and the results follow.
But I haven't used "$\epsilon$" at all...I'm thinking I need to prove that $\forall \epsilon >0$, $\exists N$ s.t whenever $n>N, |a_n-\lim_{{n}\to{\infty}}b_n|=\epsilon$.

Hi Rido12, :)

You are given that, $\lim_{{n}\to{\infty}}|a_n-b_n|=0$. Hence for all $\epsilon > 0$ there exist $N_1>\mathbb{N}$ such that,

\[|a_n-b_n|<\epsilon\mbox{ whenever }n>N_1\]

Suppose, $\lim_{n\rightarrow\infty}a_n=L_1$ and $\lim_{n\rightarrow\infty}b_n=L_2$. Hence for all $\epsilon > 0$ there exist $N_2>\mathbb{N}$ such that,

\[|a_n-L_1|<\epsilon\mbox{ whenever }n>N_2\]

and for all $\epsilon > 0$ there exist $N_3>\mathbb{N}$ such that,

\[|b_n-L_2|<\epsilon\mbox{ whenever }n>N_3\]

Adding the above two inequalities and using the triangle inequality we get,

\[|(a_n-b_n)-(L_1-L_2)|<2\epsilon\mbox{ whenever }n>\mbox{Max }(N_2,N_3)\]

Adding the first inequality to the above result and again using the triangle inequality we get,

\[|L_1-L_2|<3\epsilon\mbox{ whenever }n>\mbox{Max }(N_1, N_2, N_3)\]

Hence,

\[L_1=L_2\]

 

[Dethrone]

Hi Rido12, :)

You are given that, $\lim_{{n}\to{\infty}}|a_n-b_n|=0$. Hence for all $\epsilon > 0$ there exist $N_1>\mathbb{N}$ such that,

\[|a_n-b_n|<\epsilon\mbox{ whenever }n>N_1\]

Suppose, $\lim_{n\rightarrow\infty}a_n=L_1$ and $\lim_{n\rightarrow\infty}b_n=L_2$. Hence for all $\epsilon > 0$ there exist $N_2>\mathbb{N}$ such that,

\[|a_n-L_1|<\epsilon\mbox{ whenever }n>N_2\]

and for all $\epsilon > 0$ there exist $N_3>\mathbb{N}$ such that,

\[|b_n-L_2|<\epsilon\mbox{ whenever }n>N_3\]

Adding the above two inequalities and using the triangle inequality we get,

\[|(a_n-b_n)-(L_1-L_2)|<2\epsilon\mbox{ whenever }n>\mbox{Max }(N_2,N_3)\]

Adding the first inequality to the above result and again using the triangle inequality we get,

\[|L_1-L_2|<3\epsilon\mbox{ whenever }n>\mbox{Max }(N_1, N_2, N_3)\]

Hence,

\[L_1=L_2\]

Oh wow, that makes a lot of sense, thanks! Just one clarification, so at the end, we can pick $\epsilon^*=3\epsilon$ and so it follows the standard form that whenever $n>\mbox{Max }(N_1, N_2, N_3)$, $|L_1-L_2|<\epsilon^*$? Should we be concerned that $\epsilon^*>\epsilon$, or does it not matter because it is arbitrary?

 

[Opalg]

Let ${a_n}$ and ${b_n}$ be two convergent sequences such that $\lim_{{n}\to{\infty}}|a_n-b_n|=0$. Give an $\epsilon-N$ proof to show that $\lim_{{n}\to{\infty}}a_n=\lim_{{n}\to{\infty}}b_n$.

Right now, I've just said without loss of generality, assume that $a_n>b_n$ for all $n>N$, then since $\lim_{{n}\to{\infty}}|a_n-b_n|=0$, we have $\lim_{{n}\to{\infty}}a_n-\lim_{{n}\to{\infty}}b_n=0$ and the results follow.
But I haven't used "$\epsilon$" at all...I'm thinking I need to prove that $\forall \epsilon >0$, $\exists N$ s.t whenever $n>N, |a_n-\lim_{{n}\to{\infty}}b_n|=\epsilon$.

Actually, that is a very serious loss of generality. There is no reason to think that one of the sequences will always be greater than the other. For example, you might have $a_n = \dfrac{(-1)^n}n$ and $b_n = \dfrac{(-1)^{n-1}}n$. In that example, both sequences converge to $0$, and both sequences alternate in sign. But whenever one of them is positive the other one is negative. So you cannot say that either of them is always greater than the other.

To prove the result, you need to use some argument like the proof that Sudharaka suggests.

 

[Sudharaka]

Oh wow, that makes a lot of sense, thanks! Just one clarification, so at the end, we can pick $\epsilon^*=3\epsilon$ and so it follows the standard form that whenever $n>\mbox{Max }(N_1, N_2, N_3)$, $|L_1-L_2|<\epsilon^*$? Should we be concerned that $\epsilon^*>\epsilon$, or does it not matter because it is arbitrary?

It doesn't matter since $\epsilon$ is arbitrary. Hence $\epsilon*$ is also arbitrary. If you had selected $\frac{\epsilon}{3}$ as the arbitrary value in the beginning instead of $\epsilon$ then you would have ended up with $\epsilon$ at the end. Some people do this but I prefer to just go with $\epsilon$ since even if you end up with a multiplication of $\epsilon$ it doesn't really matter as it's again arbitrary. :)

 

[Dethrone]

It doesn't matter since $\epsilon$ is arbitrary. Hence $\epsilon*$ is also arbitrary. If you had selected $\frac{\epsilon}{3}$ as the arbitrary value in the beginning instead of $\epsilon$ then you would have ended up with $\epsilon$ at the end. Some people do this but I prefer to just go with $\epsilon$ since even if you end up with a multiplication of $\epsilon$ it doesn't really matter as it's again arbitrary. :)

Ah, yes, I'm used to seeing $\epsilon/3$. I always thought it was neat to end a proof with $\epsilon /3 + \epsilon/3+\epsilon/3=\epsilon$. I'm not too great at delta-epsilon proofs...is that a skill that gets honed with more advanced courses like real analysis (not really sure what that is actually :P)? (Wondering)

 

[Nono713]

Epsilon-delta proofs are nice as a "first principles" tool to really understand what is going on, the various conditions needed, etc.. once you prove the basic limit theorems relating to function continuity, those are much easier to use. For instance your question instantly follows from the continuity of the absolute value function.

 

[Sudharaka]

Ah, yes, I'm used to seeing $\epsilon/3$. I always thought it was neat to end a proof with $\epsilon /3 + \epsilon/3+\epsilon/3=\epsilon$. I'm not too great at delta-epsilon proofs...is that a skill that gets honed with more advanced courses like real analysis (not really sure what that is actually :P)? (Wondering)

Ha ha... I don't know. Perhaps you can improve by doing more problems of this kind or it might also come with getting a more thorough understanding about the concepts itself in which case the below link might help. Works differently for different people I guess. :p :)

https://www.khanacademy.org/math/differential-calculus/limits_topic/epsilon_delta