# How to find $A_1A_2 + \cdots + A_{2010} A_{2011}$, where $A_{n+1} = \frac{1}{1+\frac{1}{A_n}}$

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.

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Is that supposed to be $$A_{n+1}= \frac{1}{1+\frac{1}{A_n}}$$or $$A_{n+1}=\frac{1}{1} + \frac{1}{A_n}\ ?$$ –  Arturo Magidin Jun 2 '12 at 4:51
@ArturoMagidin:The first one is correct –  mgh Jun 2 '12 at 4:55
@ArturoMagidin:I am really sorry as the way I typed my question was very confusing.But I dont know how to write them the way u have wrote. –  mgh Jun 2 '12 at 4:57
@user1396721: You can find guides to using LaTeX here and here –  Zev Chonoles Jun 2 '12 at 4:59

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 $$\frac{1}{n(n+1)}=\frac{(n+1)-n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.$$

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Zev Chonoles:what is induction? –  mgh Jun 2 '12 at 5:03
Zev Chonoles::I observed that:A1=1 , A2=1/2 , A3=1/3... –  mgh Jun 2 '12 at 5:13
:thus An=1/n , An+1=1/n+1..... –  mgh Jun 2 '12 at 5:14
Because $$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ as I noted above, we have that $$A_1A_2=\frac{1}{1}\frac{1}{2}=\frac{1}{1}-\frac{1}{2}$$ $$A_2A_3=\frac{1}{2}\frac{1}{3}=\frac{1}{2}-\frac{1}{3}$$ $$\cdots$$ $$A_{2010}A_{2011}=\frac{1}{2010}\frac{1}{2011}=\frac{1}{2010}-\frac{1}{2011}$$ so that $$A_1A_2+\cdots+A_{2010}A_{2011}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}‌​+\frac{1}{3}-\cdots+\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2010}-\frac{1}{2011}$$ A lot of terms cancel :) –  Zev Chonoles Jun 2 '12 at 5:30
:Thanks a lot!!! Had great fun:) –  mgh Jun 2 '12 at 5:40

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:

1. Solve the equation $x = (\alpha{}x+\beta)/(\gamma{}x+\delta)$.
2. If the equation has two distinct roots, say, $x_1$ and $x_2$, the sequence $\big\langle(A_n-x_1)/(A_n-x_2)\big\rangle_{n>0}$ is a geometric progression(AP). Goto 4.
3. Otherwise, the equation has two same roots, say, $x_0$. The sequence $\big\langle(A_n-x_0)^{-1}\big\rangle$ is an arithmetic progression(GP).
4. Find a closed-form for the AP or GP, then get the solution of the recurrence.

Some degenerate cases are not discussed, but they're trivial.

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