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 August 15th, 2014, 08:30 AM #1 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry Roots Denominators in a Series I encountered this question: $\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+ \sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1} {\sqrt{1,000,000}+\sqrt{1,000,001}}=\sqrt{a}-\sqrt{b}$ Determine the value of a and b. How to determine the denominator in that series? On a side note, why are there 2 squareroot codes which can't be displayed properly? Last edited by greg1313; August 15th, 2014 at 05:05 PM. Reason: Fixed square root symbols
 August 15th, 2014, 09:08 AM #2 Senior Member     Joined: Feb 2010 Posts: 701 Thanks: 136 Your latex code showed \sqr t. There should be no space. To simplify, rewrite each denominator so that the larger radical is first. Multiply each fraction by its conjugate. Then telescope the sum.
August 15th, 2014, 09:22 AM   #3
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Quote:
 Originally Posted by mrtwhs Your latex code showed \sqr t. There should be no space. To simplify, rewrite each denominator so that the larger radical is first. Multiply each fraction by its conjugate. Then telescope the sum.
I am eating right now, so I can't calculate on paper in the current time. However I can imagine that all denominators will be 1, right? Due to $(a-b)^2=(a+b)(a-b)$ formula.

 August 15th, 2014, 04:34 PM #4 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry I have been imagining to solve this question before I went to sleep last night, and the answer is $\sqrt{1,000,001}-\sqrt{2}$, right?
 August 15th, 2014, 04:55 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,557 Thanks: 2558 Math Focus: Mainly analysis and algebra Yes. well done. I think it's a better (more elegant) question as $$\sum_{k=1}^{n^2 - 1}\frac{1}{\sqrt{k} + \sqrt{k+1}}$$
 August 16th, 2014, 02:34 AM #6 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry OK, now on a side note, why weren't the roots displayed properly last night despite I had been checking and I didn't write "\sqr t" but normal "\sqrt"?

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