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
    Commented Dec 26, 2015 at 17:09
  • 2
    $\begingroup$ Do you mean $b<a$ in your assumption for a contradiction? $\endgroup$
    – Micapps
    Commented Dec 26, 2015 at 17:10
  • $\begingroup$ @Micapps Yes, that was a typo $\endgroup$
    – macurie
    Commented Dec 26, 2015 at 17:55

1 Answer 1


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
    Commented Dec 27, 2015 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
    Commented Dec 27, 2015 at 0:58
  • $\begingroup$ So your proof is a direct proof as you don't assume $a>b$? $\endgroup$
    – macurie
    Commented Dec 27, 2015 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
    Commented Dec 27, 2015 at 14:46

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