# How does $x_{n+1}\ge x_n\land y_{y+1}\le y_n \land \lim_{n\to\infty}(x_n-y_n) = 0\implies \lim x_n = \lim y_n$ and both $x_n$ and $y_n$ are convergent

Let $$x_n$$ and $$y_n$$ denote two sequences. The sequences are given such that: $$x_{n+1} \ge x_n \\ y_{n+1} \le y_n \\ \lim_{n\to\infty}(x_n-y_n) = 0$$ Prove that both $$x_n$$ and $$y_n$$ are convergent and: $$\lim_{n\to\infty} x_n = \lim_{n\to\infty}y_n$$

I'm not sure how to proceed with the proof. I've started with inspecting the relations between $$x_n$$ and $$y_n$$: $$x_{n+1} \ge x_n \\ y_{n+1} \le y_n \\ \iff \\ x_{n+1} \ge x_n \\ -y_{n+1} \ge -y_n$$ From which it follows that: $$x_{n+1} - y_{n+1} \ge x_n - y_n \tag 1$$

Denote $$z_n = x_n - y_n$$, then by $$(1)$$ it follows that $$z_n$$ is a monotonically increasing sequence: $$z_{n+1} \ge z_n$$ By monotone convergence theorem it follows that if $$\lim z_n = 0$$ and $$z_{n+1} \ge z_n$$ then $$z_n \le 0$$. Also $$z_n$$ is convergent hence bounded: $$m \le z_n \le M$$ By the fact that $$z_n < 0$$ it follows that $$x_n < y_n$$. The problem is i do not see how to combine those facts in order to show what is requested in the problem statement. Seems like the problem may be reduced to using MCT for $$x_n$$ and $$y_n$$ alone or use the properties of $$\limsup$$, $$\liminf$$ or somehow use the triangular inequality but i do not see how.

What is a proper way to achieve that?

• Hint: Can you use $x_n<y_n$ to show that $\{x_n\}$ is bounded? – SmileyCraft Jan 10 '19 at 14:04
• @Uncountable $\lim(x_n - y_n) = \lim (n - (-n)) \ne 0$ so it violates the initial conditions. – roman Jan 10 '19 at 14:05
• Yes, roman, I realised that as soon as I posted it. I apologise. – Uncountable Jan 10 '19 at 14:07
• @SmileyCraft I would appreciate if you could elaborate on this. Intuitively the statement is true however I don't see where the boundedness of $x_n$ comes from – roman Jan 10 '19 at 14:13
• @roman By induction $y_n\leq y_1$, so $y_1$ is an upper bound of $\{x_n\}$. – SmileyCraft Jan 10 '19 at 14:15

I would proceed as follows, without reference to the “nested intervals lemma” (which is essentially equivalent to what you want to prove):

As you already observed, $$z_n = x_n - y_n$$ is increasing, with $$\lim_{n \to \infty} z_n = 0$$. It follows that $$x_n \le y_n$$ for all $$n \in \Bbb N$$.

Then $$x_n \le x_{n+1} \le y_{n+1} \le y_1 \quad (n \in \Bbb N)$$ so that the sequence $$(x_n)$$ is increasing and bounded above, and therefore convergent: $$\lim_{n \to \infty} x_n$$ exists.

Similarly, $$y_n \ge y_{n+1} \ge x_{n+1} \ge x_1 \quad (n \in \Bbb N)$$ implies that $$(y_n)$$ is decreasing and bounded below, so that $$\lim_{n \to \infty} y_n$$ exists as well.

Finally (since both limits exits), $$0 = \lim_{n\to\infty}(x_n-y_n) = \lim_{n \to \infty} x_n - \lim_{n \to \infty} y_n$$ so that the limits are equal.

• Nice approach, thank you! – roman Jan 10 '19 at 19:39

Using a hint by @SmileyCraft we might show that $$x_n$$ is bounded and increasing hence convergent. Consider the following inequality: $$x_n \le y_n$$ Since $$x_n$$ is monotonically decreasing we have that: $$x_n \le x_{n+1}$$

By $$y_n$$ is monotonically decreasing: $$y_{n+1} \le y_{n}$$

Expanding that further and using the fact $$x_n \le y_n$$ we may obtain: $$x_1 \le x_2 \le \cdots \le x_{n-1} \le x_{n} \le x_{n+1} \le y_{n+1} \le y_n \le y_{n-1} \le \cdots \le y_2 \le y_1$$

Which a set of nested intervals. By the Nested intervals lemma and the fact that $$\lim_{n\to\infty}(x_n - y_n) = 0$$ we know that there is a single point $$L$$ which belongs to all of the intervals. By monotonicity of $$x_n$$ and $$y_n$$ this point appears to be the limit for both sequences. So by this $$x_n$$ and $$y_n$$ converge and: $$\lim_{n\to\infty} x_n = \lim_{n\to\infty} y_n =L$$

• I hope this is correct – roman Jan 10 '19 at 14:53