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If a sequence ($a_n$) is monotonically increasing, and ($b_n$) is a decreasing sequence, with $\lim_{n\to\infty}\,(b_n-a_n)=0$, show that $\lim a_n$ and $\lim b_n$ both exist, and that $\lim a_n=\lim b_n$.

My attempt:

To show that the limits of both sequences exist, I think I should be using the Monotone Convergence Theorem (MCT). For that I would need to show that the sequences are bounded.

($a_n$) is increasing, and so it should be bounded below. ($b_n$) is decreasing, so it should be bounded above. The challenge here is to show that ($a_n$) can be bounded above and ($b_n$) can be bounded below. This should utilise the third condition, from which I get:

$$\begin{align*} & \lim_{n\to\infty}\,(b_n-a_n)=0 \\[3pt] \iff & \forall\varepsilon>0,\ \exists N\in \mathbb{N} \text{ s.t. } \forall n\geq N,\ |{b_n-a_n}|<\varepsilon \end{align*}$$

I then tried using the triangle inequality: $$ |b_n|-|a_n|\leq|b_n-a_n|<\varepsilon$$

but I'm not sure where to go from here.

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    $\begingroup$ (+1) for the fluffy little dog (and a well posed question). $\endgroup$ – tired Feb 4 '17 at 9:41
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    $\begingroup$ @tired It's my favourite dog breed. Thanks, haha. :) $\endgroup$ – Troy Feb 4 '17 at 10:05
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The leftmost part of the inequality $|b_n|-|a_n|\leq|b_n-a_n|<\epsilon$ isn't needed in the proof.

Claim 1     $(a_n)$ is bounded above.

Proof    Let $\epsilon>0$. From $|b_{n+1}-a_{n+1}|<\epsilon$, $a_{n+1}<b_{n+1}+\epsilon$. Use the monotonicity of $(a_n)$ and $(b_n)$. We have $$a_1 \le \dots \le a_n\le a_{n+1} < b_{n+1}+\epsilon \le b_n+\epsilon \le \dots \le b_1 + \epsilon.$$ Since the choice of $n$ in the above inequality is arbitrary, we have $a_n < b_1 + \epsilon$ for all $n \in \Bbb N$. Therefore, $(a_n)$ is bounded above by $b_1 + \epsilon$.

Similarly, we have another claim.

Claim 2     $(b_n)$ is bounded below.

Now, recall that $(a_n)$ and $(b_n)$ are increasing and decreasing sequences respectively, and apply MCT to $(a_n)$ and $(b_n)$ to establish the existence of $\lim a_n$ and $\lim b_n$. Finally, use $\lim\limits_{n\to+\infty}(b_n-a_n)=0$ to conclude that $\lim a_n = \lim b_n$.


Sorry for using others' ideas in my solution. I would like to draw a commutative diagrams in the comments, but the system forbids me from posting comments with two or more @ characters, so I can't post the following diagram in a comment. Hoping that others can benefit from his answer at the first glance, I draw this diagram for fun.

A graphical explanation to DonAntonio's answer

$\require{AMScd}$ \begin{CD} @. a_n \\ @. @AA \vdots A \\ @. a_{N+1} \\ @. @AA (a_n)\uparrow A \\ @. a_N \\ \text{Suppose }a_N > b_K. \\ b_K @. \\ @V(b_n)\downarrow VV @.\\ b_{K+1} @. \\ @V \vdots VV @.\\ b_n @. \end{CD} It's clear from the diagram that we have to take $n \ge \max\{K,N\}$. But $\lim\limits_{n\to+\infty}(b_n-a_n)=0$, contradiction.

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    $\begingroup$ Nice explanation to my answer. +1 $\endgroup$ – DonAntonio Feb 4 '17 at 12:57
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Hint:

Suppose there exists $\;K\in\Bbb N\;$ such that for some $\;N\in\Bbb N\;,\;\;a_N>b_K\;$ , say $\;a_N-b_K=\epsilon>0\;$ , but then

$$\forall\,n\ge N\,,\,\,\forall\,m\ge K\;,\;\;\begin{cases}a_n\ge a_N>b_K,&\text{since $\,\{a_n\}\,$ is monotone ascending}\\{}\\ b_m\le b_K<a_N,&\text{since $\,\{b_n\}\,$ is monotone descending}\end{cases}$$

and from here we'd get that for any $\;n\ge\max\,\{K,N\}\;$ :

$$a_n\ge a_N>b_K\ge b_n\implies a_n-b_n>\epsilon>0\implies \lim_{n\to\infty}(a_n-b_n)\neq0$$

and the above shows not only both sequence are bounded in the right direction, but also that they both bound each other resp.

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  • $\begingroup$ If I employ a graphical argument, if I have an increasing sequence $(a_n)$ and decreasing sequence $(b_n)$ and they have the same limit, I don't see how and when $a_N>b_K$. $\endgroup$ – Troy Feb 4 '17 at 10:04
  • $\begingroup$ @Troy Exactly the answer's point ! It can't be, and thus the whole sequence $\;a_n\;$ is bounded above by the $\;b_n\,$'s . Didn't you understand what the above is trying to tell you? $\endgroup$ – DonAntonio Feb 4 '17 at 10:22
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    $\begingroup$ I did not, but now I do. Thanks for taking the time to explain your answer! $\endgroup$ – Troy Feb 4 '17 at 10:33
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    $\begingroup$ @Troy No worries. Hoping that you can benefit from this answer, I attempted to post a commutative diagram in a comment, but the system doesn't let me post it due to the presence of more than one @ character, so I have to post it in my answer. Btw, +1 $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 4 '17 at 11:12
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You choose an $\varepsilon>0$ and have $$|(a_n-b_n)|\lt\varepsilon, \forall n> N_0$$ because $\lim (a_n-b_n)=0$. To finish your proof, assume that $\lim a_n=a$ and $\lim b_n=b$. Then you have $$|a_n-a|<\varepsilon, \;\forall n>N_1$$ and $$|b_n-b|<\varepsilon, \; \forall n>N_2$$ So select an $n$ such that $n>N_0, n>N_1, n>N_2$ to get $$|a-b|\le|a-a_n|+|a_n-b_n|+|b_n-b|\lt 3\varepsilon$$

$\varepsilon$ was chosen arbitrary therefore $$|a-b|=0$$

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Since $\lim_{n\to\infty}(b_n-a_n)=0$, there is an $N$ such that $|a_n-b_n|<1$ for all $n\ge N$. ($1$ is a number that I have just chosen for $\varepsilon$.) Since $b_n$ is decreasing, we have $a_n<b_n+1\le b_N+1$ for all $n\ge N$. Therefore $a$ is bounded from above, indeed by $\max\left\{a_0,a_1,\dots,a_{N-1},b_N+1\right\}$.

Therefore $a$ converges. Since $b_n = a_n + (b_n-a_n)$ and both $a$ and $b-a$ converge, $b$ also converges.

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