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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    $\begingroup$ Is that supposed to be $$A_{n+1}= \frac{1}{1+\frac{1}{A_n}}$$or $$A_{n+1}=\frac{1}{1} + \frac{1}{A_n}\ ?$$ $\endgroup$ – Arturo Magidin Jun 2 '12 at 4:51
  • $\begingroup$ Welcome to math.SE. In order to get the best possible answers, it is helpful if you say what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Also, please provide an answer to Arturo's question about your intended notation - I chose what I assumed you meant, but you need to specify the correct one. $\endgroup$ – Zev Chonoles Jun 2 '12 at 4:52
  • $\begingroup$ @ArturoMagidin:The first one is correct $\endgroup$ – mgh Jun 2 '12 at 4:55
  • $\begingroup$ @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. $\endgroup$ – mgh Jun 2 '12 at 4:57
  • $\begingroup$ @user1396721: You can find guides to using LaTeX here and here $\endgroup$ – 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 $$\frac{1}{n(n+1)}=\frac{(n+1)-n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.$$

  • $\begingroup$ Zev Chonoles:what is induction? $\endgroup$ – mgh Jun 2 '12 at 5:03
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    $\begingroup$ Zev Chonoles::I observed that:A1=1 , A2=1/2 , A3=1/3... $\endgroup$ – mgh Jun 2 '12 at 5:13
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    $\begingroup$ :thus An=1/n , An+1=1/n+1..... $\endgroup$ – mgh Jun 2 '12 at 5:14
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    $\begingroup$ 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 :) $\endgroup$ – Zev Chonoles Jun 2 '12 at 5:30
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    $\begingroup$ :Thanks a lot!!! Had great fun:) $\endgroup$ – 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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