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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 something – anonymous Nov 15 '10 at 19:56
@DEbanjan: Some condition on $a_{1},a_{2}, \cdots$ should be there i suppose – anonymous Nov 15 '10 at 19:56
@Chandru: Yes. a_i are in arithmetic progression, I believe. – Aryabhata Nov 15 '10 at 19:57
@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. – Aryabhata Nov 15 '10 at 20:01
@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. – Arturo Magidin Nov 15 '10 at 20:17
up vote 5 down vote accepted

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 $m<n$. Canceling the sum $$ \frac{1}{a_1a_n}+\frac{1}{a_na_1}$$ from the left with the term $$\frac{2}{a_1+a_n}\left(\frac{1}{a_1}+\frac{1}{a_n}\right)$$ on the right and using the induction hypothesis on the remaining terms $$\frac{1}{a_2a_{n-1}}+\cdots +\frac{1}{a_{n-1}a_2},$$ we find that we need to prove $$\frac{2}{a_2+a_{n-1}}$$ on the left is equal to $$\frac{2}{a_1+a_{n}}$$ on the right. And this is true because $a_2+a_{n-1}=a_1+a_n=2a_1+(n-1)d$.

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