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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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Hint: Try rationalizing the denominators.

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    $\begingroup$ @DavidH I've been waiting for a chance to use it. Haha. $\endgroup$ Jun 23 '14 at 3:16
  • $\begingroup$ i did that.. thank you for your hint..@CameronWilliams $\endgroup$
    – David
    Jun 23 '14 at 3:49
  • $\begingroup$ You're very welcome @David $\endgroup$ Jun 23 '14 at 5:06
  • $\begingroup$ The link appears, thus the whole answer, to be broken. $\endgroup$
    – Hanno
    Jan 10 '21 at 11:22
  • $\begingroup$ ... or not accessible $\:\ddot\frown$ $\endgroup$
    – Hanno
    Jan 10 '21 at 11:28
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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 cancelled. 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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