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August 15th, 2014, 07:30 AM  #1 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,961 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 04:05 PM. Reason: Fixed square root symbols 
August 15th, 2014, 08:08 AM  #2 
Senior Member Joined: Feb 2010 Posts: 679 Thanks: 127 
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, 08:22 AM  #3 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,961 Thanks: 132 Math Focus: Trigonometry  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 $(ab)^2=(a+b)(ab)$ formula.

August 15th, 2014, 03:34 PM  #4 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,961 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, 03:55 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 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, 01:34 AM  #6 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,961 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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