# induction proof

• Dec 19th 2006, 01:00 PM
Swamifez
induction proof
I've been working on this for the past hour, but haven't gone anywhere with it. If anyone can help to complete it, it would be highly appreciated. Thanks

Let 0< a1< b1 and define

an+1= √anbn

bn+1=(an+bn)/2

a) Use induction to show that
an<an+1<bn+1<bn

Thus prove that an and bn converge.
b) Prove that they have the same limit.
• Dec 19th 2006, 01:31 PM
Plato
How far along the 'path' have you gotten?
Please tell us where you are in the proof.
• Dec 19th 2006, 06:26 PM
Swamifez
Greatly appreciate it if someone took me through the entire problem, don't know where to start. Thank you
• Dec 20th 2006, 06:03 AM
Plato
This entire problem revolves about the following facts:
$\begin{array}{l}
x^2 + y^2 \ge 2xy \\
0 < a_N < b_N \quad \Rightarrow \quad a_N < \frac{{a_N + b_N }}{2} < b_N \\
\end{array}.$

Begin with this, if $0 < a_N < b_N$ we can use the above and derive:
$\begin{array}{l}
a_{N + 1} = \sqrt {a_N b_N } > \sqrt {a_N a_N } = a_N \\
a_N < \frac{{a_N + b_N }}{2} < b_N \quad \Rightarrow \quad a_N < b_{N + 1} < b_N \\
b_{N + 1} = \frac{{a_N + b_N }}{2} \ge \sqrt {a_N b_N } = a_{N + 1} \\
\end{array}.$
• Dec 21st 2006, 06:19 AM
CaptainBlack
Quote:

Originally Posted by Swamifez
Can someone just show me a full blown proof of this problem, I really would have posted everything I had, but its really hard for me to type everything I had. Thanks

Let 0< a1< b1 and define

an+1= √anbn

bn+1=(an+bn)/2

a) Use induction to show that
an<an+1<bn+1<bn

Suppose that for some $k \in \mathbb{N}$ that:

$0

then $a_{k+1}=\sqrt{a_kb_k}>a_k$ as $b_k>a_k$,

and $b_{k+1}=(a_k+b_k)/2 as $a_k.

Also by the geometric-arithmetic mean inequality:

$a_{k+1} (equality is not a possibility here as $a_k \ne b_k$),

so we have shown that:

$0.

The base case is demonstrated by showing that:

$0

using essentialy the same methods, which proves the required result by
mathematical induction.

RonL
• Dec 21st 2006, 06:25 AM
CaptainBlack
Quote:

Originally Posted by Swamifez
Let 0< a1< b1 and define

an+1= √anbn

bn+1=(an+bn)/2

a) Use induction to show that
an<an+1<bn+1<bn

Thus prove that an and bn converge.
b) Prove that they have the same limit.

Part a) shows that $a_n$ is an increasing bounded sequence, hence converges,
also $b_n$ is a decreasing sequence bounded below and hence converges.

The limits satisfy the equations:

$a=\sqrt{a\,b}$

$b=(a+b)/2$.

The solution to these is $a=b$.

RonL