My question is:
If $A_{n+1} = \frac{1}{1+\frac{1}{A_n}}$ ($n\in\mathbb{N}$) and $A_1=1$, then find the value of: $A_1A_2 + A_2A_3 + A_3A_4 + \cdots + A_{2010} A_{2011}.$
Please I would like to get some hints to solve this question.
My question is:
If $A_{n+1} = \frac{1}{1+\frac{1}{A_n}}$ ($n\in\mathbb{N}$) and $A_1=1$, then find the value of: $A_1A_2 + A_2A_3 + A_3A_4 + \cdots + A_{2010} A_{2011}.$
Please I would like to get some hints to solve this question.
Here is a hint: Calculate the first few values of $A_n$; you will notice a clear pattern which you can prove to be true in general with induction. Then, note that $\frac{1}{n(n+1)}=\frac{(n+1)-n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.$
Notice that $A_{n+1} = (1+A_n^{-1})^{-1} = A_n/(1+A_n)$, we get $A_{n+1}^{-1} = 1 + A_n^{-1}$, and the recurrence relation $A_{n+1} = (\alpha{}A_n+\beta)/(\gamma{}A_n+\delta)$ where $\gamma\ne0$ can be solved systematically:
Some degenerate cases are not discussed, but they're trivial.