Sequence

**Nichelle14** · July 5th 2006, 03:59 PM

How do I do this problem?

Is the sequence An = sqrt(n^2+n+1) -sqrt(n^2-n+1) convergent? Find it's limit.

**Soroban** · July 5th 2006, 04:25 PM

Hello, Nichelle14!

Is the sequence: $A_n \:= \:\sqrt{n^2+n+1} - \sqrt{n^2-n+1}$ convergent? .Find its limit.

Multiply top and bottom by the conjugate:

. . $\frac{\sqrt{n^2+n+1} - \sqrt{n^2-n+1}}{1}\cdot\frac{\sqrt{n^2+n+1} + \sqrt{n^2 - n+1}}{\sqrt{n^2+n+1} + \sqrt{n^2-n+1}}$

. . $= \;\frac{(n^2+n+1) - (n^2-n+1)}{\sqrt{n^2+n+1} + \sqrt{n^2-n+1}} \;= \;\frac{2n}{\sqrt{n^2+n+1} + \sqrt{n^2-n+1}}$

Divide the top by $n$ ; divide the bottom by $\sqrt{n^2} = n$

. . $\frac{\frac{2n}{n}}{\frac{\sqrt{n^2+n+1}}{\sqrt{n^ 2}} + \frac{\sqrt{n^2-n+1}}{\sqrt{n^2}}} \;=\;\frac{2}{\sqrt{\frac{n^2}{n^2} + \frac{n}{n^2} + \frac{1}{n^2}} + \sqrt{\frac{n^2}{n^2} - \frac{n}{n^2} + \frac{1}{n^2}}}$

. . $=\;\frac{2}{\sqrt{1 + \frac{1}{n} + \frac{1}{n^2}} + \sqrt{1 - \frac{1}{n} + \frac{1}{n^2}}}$

Then: . $\lim_{n\to\infty}\left[\frac{2}{\sqrt{1 + \frac{1}{n} + \frac{1}{n^2}} + \sqrt{1 - \frac{1}{n} + \frac{1}{n^2}}}\right]\;=$ $\frac{2}{\sqrt{1+0+0} + \sqrt{1 + 0 + 0}}\;=\;\frac{2}{1+1}$

Therefore: . $\boxed{\lim_{n\to\infty}\left(\sqrt{n^2+n+1} - \sqrt{n^2-n+1}\right)\;=\;1\,}$

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