9
$\begingroup$

I need some help with understanding a part of this proof and also writing it up correctly. Given $a_n\geq a_{n+1}\geq b_{n+1} \geq b_n$ with $a_1=a$ and $b_1=b$. I am also given that $$a_{n+1}=\frac{a_n+b_n}{2}$$ and $$b_{n+1}=\sqrt{a_nb_n}$$ I need to show that sequences ${a_n}$ and ${b_n}$ converges and that ${a_n}$ and ${b_n}$ have the same limit.

I am told to use the monotonic convergence theorem to prove that both sequences converges and I have the following proof:

Notice that {$a_n$} is monotonically decreasing while {$b_n$} is monotonically increasing. Since {$a_n$} is bounded above by supremum $a_1$ below by its infimum $b_1$, {$a_n$} according to the monotonic convergence theorem has to converge.

Similarly, notice that {$b_n$} is bounded below by infimum $b$ and supremum $a$. By monotonic convergence theorem {$b_n$} must also converge as well.

Next, I am told to show that {$a_n$} and {$b_n$} have the same limit. In other words, if [$a_n-b_n$] as n tends to infinity must be 0. For this part, it seems to be the case that one can prove it by just showing that $a_{n+1} - b_{n+1} \leq (1/2) (a_n - b_n) $. And I know you can just show this by using the definition of the arithmetic mean, which is $a_{n+1} - b_{n+1} \leq a_{n+1} - b_n = (1/2) (a_n - b_n)$. Why is that? It seems incompletely and not so obvious to me. An explanation here would help.

Please help me edit my proof (what I have already) and clarify my understanding

$\endgroup$
4
  • 1
    $\begingroup$ $a_n = 1$, $b_n = 0$. That satisfies your requirement and they don't have the same limit. $\endgroup$
    – genisage
    Sep 15, 2014 at 19:25
  • $\begingroup$ Can you please elaborate? $\endgroup$
    – cambelot
    Sep 15, 2014 at 19:28
  • $\begingroup$ He probably means $a=1$ and $b=0$, but then $a_n = 1/2^{n-1}$ for all $n$ hence going to $0$. $\endgroup$
    – ir7
    Sep 15, 2014 at 20:43
  • $\begingroup$ Related: this and this. $\endgroup$ Sep 16, 2014 at 0:24

3 Answers 3

6
$\begingroup$

Note that from your first inequality: $$a_n > a_{n+1} > b_{n+1} > b_n$$ we know that each sequence $\{ b_n \}$ and $\{ a_n \}$ is monotonic and bounded. Therefore they both converge.

Let $L = \lim_{n\to\infty} a_n$ and $M = \lim_{n\to\infty} b_n$.

Now we also know that $$L = \lim_{n\to \infty} a_{n+1} = \lim_{n\to \infty} \frac{a_n + b_n}{2} = \frac{L+M}{2}$$ and $$M = \lim_{n\to\infty} b_{n+1} = \lim_{n\to\infty} \sqrt{a_n b_n} = \sqrt{LM}.$$

Either equation leads to your solution. $$L = \frac{L+M}{2} \implies 2L = L+M \implies L=M$$ and $$M=\sqrt{LM} \implies M^2 = LM \implies M=L$$ provided $M\neq 0$ for the second equation. However, provided that we assume the initial condition $a>b\neq 0$, we have $M > 0$ by monotonicity.

$\endgroup$
1
  • $\begingroup$ Can we find these limits $\endgroup$ Feb 24, 2019 at 4:29
2
$\begingroup$

For every $\quad x\epsilon N$ $\quad a_{ n }\ge 0,{ b }_{ n }\ge 0\quad $ using well known inequetion $$\sqrt { ab } \le \frac { a+b }{ 2 } \quad ,a\ge 0,b\ge 0$$ we get $${ a }_{ n+1 }=\frac { { a }_{ n }+{ b }_{ n } }{ 2 } \ge \sqrt { { a }_{ n }{ b }_{ n } } ={ b }_{ n+1 }$$ since $b_{ n+1 }=\sqrt { { a }_{ n }{ b }_{ n } } \ge \sqrt { { b }_{ n }^{ 2 } } ={ b }_{ n }$ and $\\ { a }_{ n+1 }=\frac { { a }_{ n }+{ b }_{ n } }{ 2 } \le { a }_{ n }$ and ${ b }_{ n }\le { a }_{ n }\le { a }_{ 1 },{ a }_{ n }\ge { b }_{ n }\ge { b }_{ 1 }$ so $\left( a_{ n } \right) $ and $\quad \left( { b }_{ n } \right) $ are monoton and bounded so they have finit limits A and B.when $\\ n\rightarrow \infty $ $${ a }_{ n+1 }=\frac { { a }_{ n }+{ b }_{ n } }{ 2 } $$ we get $$A=B$$

$\endgroup$
1
  • $\begingroup$ Can we determine these limits $\endgroup$ Feb 24, 2019 at 4:30
0
$\begingroup$

Hint: Take limit in the first relation on both sides. You might want $a$ and $b$ non-negative, otherwise $b_n$ might not be well defined as a real number.

EDIT: Also, as $a_1=a\geq b_1=b\geq 0$ and $a_{n+1}-b_{n+1}\leq 1/2(a_n-b_n)$ for all $n\geq 1$, we have: $$ 0\leq a_{n+1}-b_{n+1}\leq \frac{1}{2^n}(a-b),$$ leading to $$\lim_{n\to \infty} (a_n- b_n) = 0.$$

$\endgroup$
8
  • $\begingroup$ Can you elaborate a bit more? For instance, how does this $a_{n+1} - b_{n+1} \leq (1/2) (a_n - b_n) $ demonstrate that we have the same limit? I evaluated it and got that $sqrt(ab)\geq b$. I'm still not seeing what you mean. $\endgroup$
    – cambelot
    Sep 15, 2014 at 20:41
  • $\begingroup$ You don't need that. You have already proved that the limits exist and are finite. Say $x$ for sequence $a_n$ and $y$ for $b_n$. Using known limit taking properties, first relation becomes $x=(x+y)/2$, which leads to $x=y$. $\endgroup$
    – ir7
    Sep 15, 2014 at 20:46
  • $\begingroup$ Okay, let me see if I understand. By assigning x and y as the finite limits of the two sequences and evaluating a_n+1 and b_n+1 you basically concluded that they converge to the same limit. Is this right? However, what role do these inequalities in the last part play? They seem so random that I would like to understand why they are there. $\endgroup$
    – cambelot
    Sep 15, 2014 at 21:00
  • $\begingroup$ If you accept your inequality, then by induction, you can show that $a_{n+1}-b_{n+1}\leq 1/2^{n}(a_1-b_1)=1/2^{n}(a-b)$. Then take n to infinity. Your inequality requires $a>b$. $\endgroup$
    – ir7
    Sep 15, 2014 at 21:02
  • $\begingroup$ Sorry, I just realized that you are already assuming $a\geq b$. So the sequence difference is sandwiched between $0$ and $1/2^n(a-b)$. $\endgroup$
    – ir7
    Sep 15, 2014 at 21:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .