The definition I am working with is from Foundations of Mathematical Analysis by Johnsonbaugh and Pfaffenberger.

Def: Let $\{ a_n \}_{n=1}^\infty$ be a sequence of real numbers. We say that $\{ a_n \}_{n=1}^\infty$ has limit $L\in \Bbb R$ if for every $\epsilon > 0$, there exists a positive integer $N$, such that if $n \geq N$, then

$$|a_n - L|<\epsilon$$

In the proof of the theorem stating that the sum of the the two convergent sequences is convergent to the sum of their limits a simple strategy is to take the epsilon for each sequence individually and use the definition and divide the epsilon by two so that using triangle inequality you can say:

$$|(a_n + b_n)-(L+M)| \leq |a_n - L| + |b_n - M| < \epsilon/2 + \epsilon/2 = \epsilon$$

I don't understand why this is okay to do, I kind of understand the manipulation of $N$ so that $n$ holds the same inequality of the definition but something seems logically flawed in the assertion above and it is used often for a variety of limit proofs, particularly in proving basic algebra of limit proofs.

  • $\begingroup$ What seems logically flawed? In order for that estimate to hold one would usually take $n \geq\max(N_1,N_2)$, where $N_1$, $N_2$ correspond to $(a_n),(b_n)$ for the given $\epsilon/2$. $\endgroup$ – Alex Provost Aug 29 '16 at 4:45
  • $\begingroup$ Can you be more specific? Do you not understand (1) why it is allowed to use $\epsilon/2$ instead of $\epsilon$, (2) the inequality $|(a_n+ b_n) - (L+M)|$, or any other issues? $\endgroup$ – user99914 Aug 29 '16 at 4:46
  • $\begingroup$ What specifically in it do you not understand? Why $|(a_n+b_n)-(L+M)|\leq |a_n - L| + |b_n - M|$? That is just the triangle inequality. Why take $\epsilon/2$ for each? Because we can as guaranteed by the definition of limit. (if we can do it for any arbitrarily small number, then it doesn't matter what we call that number, so long as it is still again a number). Why is it helpful to? Because we wanted to show the property holds for $a_n+b_n$ and it is convenient to for arithmetic sake. $\endgroup$ – JMoravitz Aug 29 '16 at 4:47
  • $\begingroup$ Possibly relevant. $\endgroup$ – Git Gud Aug 29 '16 at 4:52
  • $\begingroup$ @ArcticChar (1) is my question. $\endgroup$ – M. Ebsim Aug 29 '16 at 5:00

By definition, if you give me a convergent sequence $(a_n)$ and any positive real number $\epsilon$, I can give you a threshold $N$ such that $a_n$ is $\epsilon$-close to its limit whenever $n \geq N$.

Therefore, if we want to prove that $(a_n) + (b_n)$ converges to the sum of the respective limits of $(a_n)$ and $(b_n)$, we must fix an arbitrary positive $\epsilon$ and find a threshold $N$ such that $a_n + b_n$ is $\epsilon$-close to that sum whenever $n \geq N$.

Now, $\epsilon$ is a fixed number. Therefore $\epsilon/2$ is another fixed positive real number. By the definition of the convergence of $(a_n)$, applied to the number $\epsilon/2$, we can find a threshold $N_1$ such that $a_n$ is $\epsilon/2$-close to its limit whenever $n \geq N_1$. Similarly, we can find a threshold $N_2$ such that $b_n$ is $\epsilon/2$-close to its limit whenever $n \geq N_2$. Taking $N = \max(N_1,N_2)$ and invoking your estimate concludes the proof.

  • $\begingroup$ Technically speaking, epsilon does not have to be divided 2, just that two fixed positive real numbers added together give another fixed positive number that holds for the definition? $\endgroup$ – M. Ebsim Aug 29 '16 at 5:53
  • 1
    $\begingroup$ @M.Ebsim You could have taken any two positive numbers $\epsilon_1$ and $\epsilon_2$ that satisfy $\epsilon_1 + \epsilon_2 \leq \epsilon$; choosing $\epsilon_1 = \epsilon_2 = \epsilon/2$ is just the simplest and most aesthetically pleasing choice. $\endgroup$ – Alex Provost Aug 29 '16 at 14:42

Assume some sequence $\{ a_n \}_{n=1}^\infty$ has a limit $L\in \Bbb R$. Then we can pick some fixed real number greater than zero which we call $\epsilon$, and assert there exists a positive integer $N$ dependent on $\epsilon$, so we could write this $N$ in functional notation as $N(\epsilon)$, such that if $n \geq N(\epsilon)$, then

$$|a_n - L|<\epsilon\tag{1}$$

In these proofs $\epsilon$ is a dummy variable that is some fixed real number greater than zero, that we don't in general need to know explicitly numerically, since we then extend the concept to all $\epsilon>0$ in the limiting process of convergence which follows since our $\epsilon$ was arbitrary, and so can be taken as small as we like. We could equivalently state the above limiting process in terms of some other number $\xi=\frac{\epsilon}{2}$. What we then have to do is play the $\epsilon$, $N(\epsilon)$ game again with $\xi$, $N(\xi)$, and find an$N(\xi)$, such that if $n \geq N(\xi)$, then

$$|a_n - L|<\xi$$

Here nothing is broken in the definition of convergence - you still have a positive real number $\xi$ that bounds the difference $|a_n - L|$ for all $n\ge N(\xi)$.

For two convergent sequences you want the analogous result to (1): $$|(a_n + b_n)-(L+M)| <\epsilon$$ so you can say $\{a_n +b_n\}_{n=1}^\infty$ has a limit $L+M\in \Bbb R$. To do this you just do the same thing that you did with one convergent sequence to the two sequences individually after splitting them up with the triangle inequality; that is given an $\epsilon_1>0$ such that for all $n\ge N(\epsilon_1)$ we have $|a_n - L|<\epsilon_1$, and given an $\epsilon_2>0$ such that for all $n\ge N(\epsilon_2)$ we have $|b_n - M|<\epsilon_2$, and we do this in such a manner that $\epsilon_1+\epsilon_2=\epsilon$. Now pick $N=\max(N(\epsilon_1),\,N(\epsilon_2))$; we do this since if $N(\epsilon_1)<N(\epsilon_2)$, then, for all $n\ge N(\epsilon_1)$ we have $|a_{n} - L|<\epsilon_1$, but if $n<N(\epsilon_2)$ we have $|b_{n} - M|>\epsilon_2$, picking $N$ solves this. So for all $n\geq N$ we have

$$|(a_n + b_n)-(L+M)| \leq |a_n - L| + |b_n - M| < \epsilon_1 + \epsilon_2 = \epsilon$$

Note there is nothing special about $\epsilon/2$ proofs, we usually pick these as it makes it easier when dealing with two convergent sequences to just half the $\epsilon$, then you have for $n\geq N(\frac{\epsilon}{2})$

$$|(a_n + b_n)-(L+M)| \leq |a_n - L| + |b_n - M| < \epsilon/2 + \epsilon/2 = \epsilon$$ which proves convergence since $\epsilon$ was arbitrary. Note how you fix your $\epsilon$ and then work around it i.e., one minute you look at $|a_n - L|<\epsilon$ for $n\ge N(\epsilon)$, then you may need to switch to $|a_n - L|<\frac{\epsilon}{2}$ for $n\ge N(\frac{\epsilon}{2})$, but the $\epsilon$ is the same in each case.

  • $\begingroup$ This honestly cleared up years of confusion for me. $\endgroup$ – Taylor Rendon Sep 2 '20 at 11:29

I think the slight abuses of notation are what's tripping you up.

What they do here is a couple of things: First, they let $\epsilon>0$ as usual, but then they define what I will denote as $\epsilon_0$ to be half of $\epsilon$. Surely, $\epsilon_0$ will still be greater than $0$, hence it is a perfectly valid choice of an 'epsilon' for each of the two sequences individually. Since those sequences are each convergent, then they have $N_a$ and $N_b$ respectively, such that $|a_n-L|<\epsilon_0$ and $|b_n-M|<\epsilon_o$ respectively. Moreover, from the definition of limit, we note that those inequalities hold for all $n>N_a$ for $a_n$ and all $n>N_b$ for $b_n$, hence those inequalities hold for both of them for all $n>N=\operatorname{max}\{N_a,N_b\}$.

Therefore, we have that $|a_n-L|+|b_n-M|<2\epsilon_0=\epsilon$ holding for all $n>N$ as required, and the first inequality, $|(a_n+b_n)-(L+M)|\leq|a_n-L|+|b_n-M|$ comes from the triangle inequality.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.