Let $\{a_n\}$ and $\{b_n\}$ be convergent real sequences. Assume that there exists a $N\in\mathbb{N}$ so

$a_n\le b_n$ (eq. 1)

for all $n\ge N$. Then

$\lim_{n\to\infty}a_n\le \lim_{n\to\infty}b_n$.

My attempt at proving

Let $a=\lim_{n\to\infty}a_n$ and $b=\lim_{n\to\infty}b_n$. $a$ and $b$ are real numbers because the limit of a sequence of real numbers is a real number (which is a corollary in my textbook). For proof by contradiction, assume


Let $\epsilon=\frac{a-b}{2}$. Then we can find $N_a,N_b\in\mathbb{N}$ so

$|a-a_n|\le\frac{\epsilon}{2}$ when $n\ge N_a$, (eq. 2)

$|b-b_n|\le\frac{\epsilon}{2}$ when $n\ge N_b$. (eq. 3)

In a textbook I found that I should do the following:

Choose a $n$ so $n\ge\max(N,N_a,N_b$), then eqs. (1-3) are satisfied. Then

$a-\epsilon\le a_n+\frac{\epsilon}{2}-\epsilon$

How do I get that last thing? Is it necessary to do a proof by contradiction or can one do a direct proof?

  • $\begingroup$ A contradiction works, but you should check and see whether what you assumed is in fact the correct thing to assume for a proof by contradiction $\endgroup$ – TomGrubb Dec 26 '15 at 17:09
  • 2
    $\begingroup$ Do you mean $b<a$ in your assumption for a contradiction? $\endgroup$ – Micapps Dec 26 '15 at 17:10
  • $\begingroup$ @Micapps Yes, that was a typo $\endgroup$ – macurie Dec 26 '15 at 17:55

You're almost there! I would just use $\epsilon$ instead of $\frac{\epsilon}{2}$ in eq./ineq. 2 and 3. With this proofs, it is useful to imagine the situation visually. Also, I think that in this case, a direct proof is easier.

Try to picture two reals $a$, $b$ with $a < b$ on the real line. For every $\epsilon > 0$ we choose, we have two sequences whose tails are contained in the intervals $[a-\epsilon, a+\epsilon], [b-\epsilon, b+\epsilon]$, respectively. By the tail, I mean the part of the sequence with $n \geq N$, for some big enough number $N$ that is dependent on $\epsilon$ (one could argue that writing $N(\epsilon)$ is clearer but this is seldom written this way in practice).

Now visualize the intervals $[a-\epsilon, a+\epsilon], [b-\epsilon, b+\epsilon]$ for $\epsilon$ small enough. That is, small enough that the intervals do not overlap. It is quite easy to see that taking $\epsilon$ smaller than half the distance between a and b suffices. Now, we intuitively see that all the $a_n$ are less or equal than the $b_n$ for $n \geq N$. If we take $\epsilon = \frac{b - a}{2}$ the intervals 'touch' at $\frac{a + b}{2}$ and we can show $a_n \leq \frac{a+b}{2} \leq b_n$ for $n \geq N$.

You just have to work this out formally:

For all $ n \geq N $ we have

$a_n - a \leq |a_n - a| \leq \epsilon = \frac{b - a}{2}$ so $ a_n \leq a + \frac{b - a}{2} = \frac{a + b}{2} $


$b - b_n \leq |b_n - b| \leq \epsilon = \frac{b - a}{2}$ so $ b - \frac{a - b}{2} = \frac{a + b}{2} \leq b_n $

It follows that $a_n \leq \frac{a + b}{2} \leq b_n\ \forall n \geq N$.

  • 1
    $\begingroup$ $\epsilon=\frac{a-b}{2}$ not $\frac{a+b}{2}$. Also you write $|a_n-a|\le\epsilon$ instead of $|a-a_n|\le\epsilon$. Why is that? $\endgroup$ – macurie Dec 27 '15 at 0:31
  • $\begingroup$ Corrected! That was kind of sloppy. I don't really think about the order, as I think of $|a_n - a|$ as the distance between $a_n$ and $a$. I think I (subconsciously) prefer this order slightly as $a_n$ is the thing that changes, so you would say $a_n$ approaches $a$ or write $\lim_{n \rightarrow \infty} a_n = a$, so I just write it down the way I think about it. $\endgroup$ – Ruben Dec 27 '15 at 0:58
  • $\begingroup$ So your proof is a direct proof as you don't assume $a>b$? $\endgroup$ – macurie Dec 27 '15 at 14:37
  • 1
    $\begingroup$ Yes! I removed the part that was confusing (I looked at your post once more and forgot you were assuming b > a and proving a contradiction). You can also make it a proof by contradiction by assuming $a > b$. Then, by the same logic, $a_n > b_n$ follows, which is a contradiction (but this kind of defeats the purpose of a proof by contradiction). $\endgroup$ – Ruben Dec 27 '15 at 14:46

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.