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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.

Zev Chonoles
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mgh
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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 02 '12 at 04:51
  • 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. – Zev Chonoles Jun 02 '12 at 04:52
  • @ArturoMagidin:The first one is correct – mgh Jun 02 '12 at 04:55
  • @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 02 '12 at 04:57
  • @user1396721: You can find guides to using LaTeX here and here – Zev Chonoles Jun 02 '12 at 04:59

2 Answers2

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

Zev Chonoles
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  • Zev Chonoles:what is induction? – mgh Jun 02 '12 at 05:03
  • @user1396721: See the Wikipedia page - it is a fundamental technique of proving mathematical statements. This problem should provide an excellent first experience with it. – Zev Chonoles Jun 02 '12 at 05:05
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    Zev Chonoles::I observed that:A1=1 , A2=1/2 , A3=1/3... – mgh Jun 02 '12 at 05:13
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    :thus An=1/n , An+1=1/n+1..... – mgh Jun 02 '12 at 05:14
  • Excellent! Can you use induction to prove that, in general, $A_n=\frac{1}{n}$? Note that this involves showing that $A_1=\frac{1}{1}$ (we know that already) and then showing that, assuming that $A_k=\frac{1}{k}$ for some $k$, then we also have $A_{k+1}=\frac{1}{k+1}$. – Zev Chonoles Jun 02 '12 at 05:15
  • :I did not get what you said.Sorry:( – mgh Jun 02 '12 at 05:20
  • Ok fine having aaumed that we have proved An=1/n what can be done next? – mgh Jun 02 '12 at 05:26
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    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 02 '12 at 05:30
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    :Thanks a lot!!! Had great fun:) – mgh Jun 02 '12 at 05:40
  • No problem, glad to help! If my answer is satisfactory, you can click the check mark beneath the arrows on the upper left of the answer to accept it. – Zev Chonoles Jun 02 '12 at 05:43
  • Good follow-through, good technique. +1 – davidlowryduda Jun 02 '12 at 09:21
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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.

Yai0Phah
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