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Find $$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\ldots+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}\right)$$

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2 Answers 2

Hint: Try rationalizing the denominators.

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6  
+1 for the not-so-subtle visual hint. :) –  David H Jun 23 at 3:16
1  
@DavidH I've been waiting for a chance to use it. Haha. –  Cameron Williams Jun 23 at 3:16
    
i did that.. thank you for your hint..@CameronWilliams –  gloom Jun 23 at 3:49
    
You're very welcome @David –  Cameron Williams Jun 23 at 5:06

From the Hint given by @cameron williams, I did the following. $$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{\sqrt{1}-\sqrt{3}}{-2}+\frac{\sqrt{3}-\sqrt{5}}{-2}+\ldots+\frac{\sqrt{2n-1}-\sqrt{2n+1}}{-2}\right)$$ Now the second and consectives gets cancelle.So, $$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{1}{-2}-\frac{\sqrt{2n+1}}{-2}\right)=\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\sqrt{n}\left(\frac{\sqrt{2+\frac{1}{n}}}{2}\right)=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$$

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