on the exercise 8.10 Apostol. (limit of sequence)

Question

The exercise states: prove that the limit of the sequence $$a_{n+2}=(a_na_{n+1})^{1/2} \ where \ a_1 \ge 0, a_2 \ge 0 $$

is $L = (a_1a_2^2)^{1/3}$

The solutin says: $$Let \ b_n = \frac{a_{n+1}}{a_n},$$ then $$b_{n+1}= 1/\sqrt{b_n} \ for \ all \ n$$ wich implies that $$b_{n+1}= b_1^{(-1/2)^n} \rightarrow 1 \ as \ n \rightarrow \infty$$

Consider $$\prod_{j=2}^{n+1}b_j = \prod_{j=1}^{n}(b_j)^{-1/2} $$

This implies that:$$(a_1^{1/2}a_2)^{-2/3}a_{n+1} = \left( \frac{1}{b_{n+1}} \right)^{2/3}$$...

I am having problems obtaining this last implication, I see that $$\prod_{j=2}^{n+1}b_j = \frac{a_{n+2}}{a_2} \ and \ \prod_{j=1}^{n}(b_j)^{-1/2}= \frac{a_{n+1}}{a_1} $$ But still I struggle.

Any help?

Answer 1

Multiply both sides of

$$\prod_{j=2}^{n+1}b_j = \prod_{j=1}^{n}b_j^{-1/2} $$

by $\prod_{j=1}^{n}b_j^{1/2}$ to get

$$b_{n+1}b_1^{1/2} \left(\prod_{j=2}^{n}b_j\right)^{3/2} = b_{n+1}b_1^{1/2} \prod_{j=2}^{n}b_j^{3/2} = 1.$$

This can be rewritten as

$$b_{n+1} (a_2/a_1)^{1/2} (a_{n+1}/a_2)^{3/2} = 1,$$

which can be manipulated to the equality you were asking about.

This solution should deal with the case where one of the $a_n$'s might be zero separately.

Answer 2

Consider squaring the LHS and RHS of the following equation:

$$\prod_{j=2}^{n+1}b_j = \prod_{j=1}^{n}b_j^{-1/2}\tag{0}$$

We obtain:

$$\prod_{j=2}^{n+1}(b_j)^2 = (b_1)^{c(n)}, c(n)=\frac{2}{3}(-1+(-1/2)^n)\tag{1}$$

$$\prod_{j=1}^{n}b_j^{-1} = (b_1)^{d(n)}, d(n)=c(n)\tag{2}$$