This is Chapter 3, Exercise 2 of Rudin's Principles.
Calculate $\lim\limits_{n \to \infty} \sqrt{n^2+n} -n$.
Hints will be appreciated.
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Hint: $$\frac{\sqrt{n^2+n}-n}{1} = \frac{\sqrt{n^2+n}-\sqrt{n^2}}{1}\times \frac{\sqrt{n^2+n}+\sqrt{n^2}}{\sqrt{n^2+n}+\sqrt{n^2}} = \cdots$$ I will expand more if needed.
Old John
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So $1/2$ is the final answer, huh? Thanks. – BSNL-IN Jun 11 '12 at 13:21
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Yes, 1/2 is correct. – Old John Jun 11 '12 at 13:54
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$\sqrt{n^2+n} - n= \frac{n}{\sqrt{n^2+n} + n} = \frac{1}{\sqrt{1+\frac 1n} + 1}$
Mohamed
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Or $\sqrt{n^2+n}=n \left(1+\frac 1n \right)^{\frac 12}=n \left(1 + \frac{1}{2n} -\frac{1}{8n^2} + o \left(\frac{1}{n^2} \right)\right)$ – Mohamed Jun 11 '12 at 13:05
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When you have a sequence that involves difference between two roots, it's often a good idea to try using the identity $(a-b)(a+b) = a^2-b^2$ to git rid of the root difference: $$ \sqrt{a} - \sqrt{b} = \left(\sqrt{a} - \sqrt{b}\right)\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}} = \frac{\left(\sqrt{a} - \sqrt{b}\right)\left(\sqrt{a} + \sqrt{b}\right)}{\sqrt{a} + \sqrt{b}} = \frac{a - b}{\sqrt{a} + \sqrt{b}} $$
Ayman Hourieh
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1@ Ayman Hourieh : Thanks. In general the identity $$(a-b) : \sum_{k=0}^{p-1} a^{p-1-k}b^k =a^p-b^p$$ where $p$ is a positive integer is used when we have limits with $\sqrt[p]{\text{expression 1}} - \sqrt[p]{\text{expression 2}}$ . – Mohamed Jun 11 '12 at 13:18
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To get an idea of what the answer might be, I tried inputting values: n=100, n=1000, n=1000000. e.g. for n = 1000, sqrt(1000000 + 1000) ~ 1000 + 1/2 because (1000 + 1/2)^2 = 1000000 + 2*1000*(1/2) + 1/4 ~ 1000000 + 1000.
If you can spot this, then the rest is easy.
Adam Rubinson
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You can also do this by inductively showing it is increasing and bounded, but their ideas are probably a better route.
toypajme
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