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?

  • 2
    $\begingroup$ Hint: Can you use $x_n<y_n$ to show that $\{x_n\}$ is bounded? $\endgroup$ – SmileyCraft Jan 10 '19 at 14:04
  • $\begingroup$ @Uncountable $\lim(x_n - y_n) = \lim (n - (-n)) \ne 0$ so it violates the initial conditions. $\endgroup$ – roman Jan 10 '19 at 14:05
  • $\begingroup$ Yes, roman, I realised that as soon as I posted it. I apologise. $\endgroup$ – Uncountable Jan 10 '19 at 14:07
  • $\begingroup$ @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 $\endgroup$ – roman Jan 10 '19 at 14:13
  • 1
    $\begingroup$ @roman By induction $y_n\leq y_1$, so $y_1$ is an upper bound of $\{x_n\}$. $\endgroup$ – 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.

  • $\begingroup$ Nice approach, thank you! $\endgroup$ – 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: $$ \lim_{n\to\infty} x_n = \lim_{n\to\infty} y_n =L $$

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

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