# Prove that the 2 sequences are nested intervals and give the element of the nested interval [duplicate]

$$a_{n+1} = \frac {2a_{n}b_{n}}{a_{n}+b_{n}}, b_{n+1}= \frac{a_{n}+b_{n}}{2}$$

where $$0 and $$n \in \mathbb{N}$$ and proven is that the sequence $$[a_{1} b_{1}], [a_{2}, b_{2}]...$$makes a nested interval and giving a number $$c \in \mathbb{R}$$ which is located in every interval of $$[a_{n} b_{n}]$$

I tried to use $$b_{n+1}= \frac{a_{n}+b_{n}}{2}$$ which is equal to $$\frac{2b_{n+1}}{b_{n}}=a_{n}$$, does it help to check the monotone of the sequences?

## 1 Answer

Note that $$a_{n+1},b_{n+1}$$ are the Harmonic and Arithmetic Mean of $$a_n, b_n$$ respectively.

Now we can use the fact here that $$AM\ge GM \ge HM$$ . This gives $$b_n\gt a_n,\forall n\in \mathbb{N}$$ (since $$b_1\gt a_1$$ given)

$$a_{n+1}-a_n=\frac{2a_nb_n}{a_n+b_n}-a_n$$

$$=\frac{a_n(b_n-a_n)}{a_n+b_n}\gt 0$$ gives $$a_{n+1}\gt a_n$$

Again $$b_n-b_{n+1}=\frac{b_n-a_n}2\gt 0$$ gives $$b_{n+1}\lt b_n$$

Thus clearly $$[a_{n+1},b_{n+1}]\subset [a_n,b_n] \forall n\in \mathbb{N}$$ and hence the intervals are nested

Oh , I forgot about the point $$c$$.Thanks to the comment by @Andrei

Still, I will prove why that is true.

Firstly $$\lim (b_n-a_n)=0$$ thus $$\lim b_n=\lim a_n$$

The common point is given by the above common limit.Let it be $$c$$

Now $$a_{n+1}b_{n+1}=a_nb_n=...=a_1b_1$$ can be easily checked and so

$$c^2=\lim (a_nb_n)=a_1b_1$$ and thus $$c=\sqrt(a_1b_1)=$$GM of $$a_1$$ and $$b_1$$

• And you can use the geometric mean of the initial numbers as $c$ Commented Jun 18, 2020 at 20:53