on a recursive sequence (exercise 8.14 Apostol).

Question

The exercise asks to prove convergence and find the limit of the sequence:$$a_n= \frac{b_{n+1}}{b_n},\text{ where } b_1=b_2 =1, b_{n+2} = b_{n} + b_{n+1}. $$

It also gives a hint: Show that $ \ b_{n+2}b_n - b_{n+1}^2 = (-1)^{n+1}$ by induction.

I am having problems proving $ \ b_{n+2}b_n - b_{n+1}^2 = (-1)^{n+1}$ by induction, could I have a helping hand?

Answer 1

Let $u_n=b_{n+2}b_n-b_{n+1}^2$. We have $u_n=\begin{vmatrix}b_n&b_{n+1}\cr b_{n+1}&b_{n+2}\end{vmatrix}$ hence $$u_{n+1}=\begin{vmatrix}b_{n+1}&b_{n+2}\cr b_{n+2}&b_{n+3}\end{vmatrix}=\begin{vmatrix}b_{n+1}&b_n+b_{n+1}\cr b_{n+2}&b_{n+1}+b_{n+2}\end{vmatrix}=\begin{vmatrix}b_{n+1}&b_{n}\cr b_{n+2}&b_{n+1}\end{vmatrix}=-u_n$$ Then it is easy.

Answer 2

$$b_{n+3}b_{n+1}-b_{n+2}^2= (b_{n+2}+b_{n+1})b_{n+1}-b_{n+2}^2 = (b_{n+1}-b_{n+2})b_{n+2} + b^2_{n+1} = (b_{n+1}-b_{n+2})(b_{n+1} + b_n) + b^2_{n+1}= 2b^2_{n+1}-b_{n+2}b_{n+1} +b_{n+1}b_n -b_{n+2}b_n = (b^2_{n+1}-b_{n+2}b_n)+b^2_{n+1}-b_{n+2}b_{n+1} +b_{n+1}b_n$$

Now note that $$b^2_{n+1}-b_{n+2}b_{n+1} +b_{n+1}b_n = b_{n+1}(b_{n+1}+b_n-b_{n+2}) = 0$$

and therefore you get that $$b_{n+3}b_{n+1}-b_{n+2}^2= (b^2_{n+1}-b_{n+2}b_n)+b^2_{n+1}-b_{n+2}b_{n+1} +b_{n+1}b_n = -(b_{n+2}b_n-b^2_{n+1}) = -(-1)^{n+1} = (-1)^{n+2}$$