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If $a_1,\ldots,a_n$ are in arithmetic progression and $a_i\gt 0$ for all $i$, then how to prove the following two identities:

$ (1)\large \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$

$(2) \large\frac{1}{a_1 \cdot a_n} + \frac{1}{a_2 \cdot a_{n-1}} + \frac{1}{a_3 \cdot a_{n-2}}+ \cdots + \frac{1}{a_n \cdot a_1} = \frac{2}{a_1 + a_n} \biggl( \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} \biggr)$

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    @Debanjan: I think you are missing something2010-11-15
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    @DEbanjan: Some condition on $a_{1},a_{2}, \cdots$ should be there i suppose2010-11-15
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    @Chandru: Yes. a_i are in arithmetic progression, I believe.2010-11-15
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    @ Chandru1: I guess you have edited the conditions which I re-edited now.2010-11-15
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    @Debanjan: What have you tried so far?2010-11-15
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    It's directly given as chug an plug in my module, I have no idea how to proof them and of-course I don't want to mug it.2010-11-15
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    @Debanjan: I just edited the title as it had some error.2010-11-15
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    @ Chandru1 : I knew it was due to some formatting problem ...which I was in the process of fixing.2010-11-15
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    @Debanjan: If you don't want to mug it, I suggest you try it first. At least edit the question with the ideas you had.2010-11-15
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    @ Moron : what else is better title for this problem :O ?2010-11-15
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    @Deb: I was talking about "Prove an Identity" or whatever you had before.2010-11-15
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    @Moron:I guess that was not me .. chandru1 edited to that while I was fixing the syntactic error. .. I am not getting much idea for this one,...,it is not much pleasure to type that whole thing here :)2010-11-15
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    @Chandru1 and @Debanjan: try not to use the title as the first half of a sentence that continues in the body; the body of the question should be self-contained.2010-11-15
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    @Arturo: The title had been edited by Debanjan now. I edited it because it was *showing some formatting error*.2010-11-15
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    @Arturo Magidin:Does this identity holds if $a_i < 0$ for any $i$ ?Else I think that conditions should be mentioned.2010-11-15
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    @Debanjan: I was trying to edit it so it was self-contained; I seemed to have missed a condition in your original title. Sorry about that.2010-11-15
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    @ Arturo Magidin :Nevermind :)2010-11-15

3 Answers 3

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This question is now old enough for some more complete answers.

For number 1:

$$\sum_{k=1}^{n-1} \frac{1}{ \sqrt{a_k}+ \sqrt{a_{k+1}}} = \sum_{k=1}^{n-1} \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$$ where $d$ is the common difference, $$ = \frac{1}{d} \left( \sqrt{a_n} - \sqrt{a_1} \right) = \frac{a_n - a_1}{d(\sqrt{a_n} + \sqrt{a_1})}$$ $$= \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}.$$

And for number 2:

$$ S = 2 \sum_{k=1}^n \frac{1}{a_k} = \left( \frac{1}{a_1} + \frac{1}{a_n} \right) + \left( \frac{1}{a_2} + \frac{1}{a_{n-1}} \right) + \cdots + \left( \frac{1}{a_n} + \frac{1}{a_1} \right)$$ $$ = \sum_{k=1}^n \frac{a_{n-k+1}+ a_k}{a_k a_{n-k+1}}.$$

Now $ a_{n-k+1}+ a_k = 2a_1 + (n-1)d = a_1 + a_n$ and so

$$ S = (a_1+a_n) \sum_{k=1}^n \frac{1}{a_k a_{n-k+1}}$$

from which the result follows.

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For the first one, use induction (or) note that $\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}$, where $d$ is the common difference between the successive terms. Now use the telescopic summation to cancel out the terms in the numerator and massage it to get the final expression on the right hand side.

For the second one, try to write each term on the Left Hand Side as a difference of two terms and proceed.

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Both identities can be proved quite easily with inductions. Let $d$ be the common difference, i.e. $d=a_{n+1}-a_n$.

  1. Use induction. So we need to prove $$ \frac{n-2}{\sqrt{a_1}+\sqrt{a_ {n-1}}}+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_ n}}=\frac{n-1}{\sqrt{a_1}+\sqrt{a_ n}}.$$ Rationalize the denominators and substitute $$a_{n-1}-a_1=(n-2)d, a_n-a_{n-1}=d, a_n-a_1=(n-1)d.$$ Upon cancellations, we find $$\sqrt{a_{n-1} }-\sqrt{a_ 1}+\sqrt{a_n }-\sqrt{a_{n-1} }=\sqrt{a_ n}-\sqrt{a_ 1}$$ on the left and the same on the right.

  2. Use induction, assuming true for $m