Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|cite|improve this question

2 Answers 2

Hint: Try rationalizing the denominators.

share|cite|improve this answer
+1 for the not-so-subtle visual hint. :) – David H Jun 23 '14 at 3:16
@DavidH I've been waiting for a chance to use it. Haha. – Cameron Williams Jun 23 '14 at 3:16
i did that.. thank you for your hint..@CameronWilliams – gloom Jun 23 '14 at 3:49
You're very welcome @David – Cameron Williams Jun 23 '14 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, $$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{1}{-2}-\frac{\sqrt{2n+1}}{-2}\right)=\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\sqrt{n}\left(\frac{\sqrt{2+\frac{1}{n}}}{2}\right)=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.