Converging Sequences and Limits: Help with Induction Proof?

[Swamifez]

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)/2a) 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.

 

[quasar987]

1° show it's true for n=1 first.

2° assume it's true for n=m-1, i.e. assume $a_{m-1}<a_{m}<b_{m}<b_{m-1}$

3° Use the part of the induction hypothesis that say $a_{m}<b_{m}$ to prove $a_{m+1}<a_m$ and $b_{m+1}>b_m$. For the part $a_{m+1}<b_{m+1}$, notice that since the a_i and b_i are positive, it is equivalent to showing that $(a_{m+1})^2<(b_{m+1})^2$, i.e. that $$a_nb_n<\frac{a_n^2+b_n^2}{4}+\frac{a_nb_n}{2}$$, etc. (think perfect square). That's enough hints. Go think for another hour.

 

[tacman]

First of all, it is important to note that the given sequence is well-defined since an+1 and bn+1 are always positive (since √an and (an+bn)/2 are both positive for positive an and bn).

a) Base case: For n=1, we have a2=√a1b1 and b2=(a1+b1)/2. Since 0<a1<b1, we can see that a2<a1<b1<b2, satisfying the given inequality.

Inductive step: Assume that an<an+1<bn+1<bn for some arbitrary k. We will now prove that this holds for k+1 as well.

For an+1, we have:

an+1=√anbn<√bnbn=bn

For bn+1, we have:

bn+1=(an+bn)/2>(an+an)/2=an

Combining these two inequalities, we get:

an<an+1<bn+1<bn

Thus, by the principle of mathematical induction, we can conclude that an<an+1<bn+1<bn for all positive integers n. This also implies that the sequences an and bn are both increasing and decreasing respectively, hence they converge.

b) To prove that an and bn have the same limit, let L be the limit of both sequences. We will show that L satisfies the following equations:

L=√(L^2)=L/2

Solving these equations, we get L=0. Therefore, both sequences converge to 0, hence they have the same limit.

This completes the proof by induction.