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I'm working on an old text (Buchanan's Limits: A Transition to Calculus) and I'm having a bit of trouble Proving that a certain sequence is within the general radius E of 2 (that is, that all the terms of the sequence lie in the interval (2 - E,2 + E), where E is a positive real number). Here's my work:

$$2-E<\sqrt{\frac{4n+1}{n}}<2+E$$ $$(2-E)^{2}<4+\frac{1}{n}<(2+E)^{2}$$ $$-4E+E^{2}<\frac{1}{n}<4E+E^{2}$$

And this is where I basically left it. I know $\frac{1}{n}$ cannot get larger than 1 if n is a natural number and I know that the right side of the inequality is always positive but I've always had a bit of trouble with quadratic inequalities. Ideally I'd be able to find a number M such that all n > M are in the interval.

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    I have reworked it and I think I have it right, although I used a step whose justification I'm not sure about: $$(2-E)^{2}<4+\frac{1}{n}<(2+E)^{2}$$ – 2017-02-27
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    Forgive the double comment. I have fat fingers. I have reworked it and I think I have it right, although I used a step whose justification I'm not sure about: $$2-E< \sqrt{4+\frac{1}{n}}<2+E$$ $$\sqrt{4+\frac{1}{n}}>2$$ True for all n $$2-E<2$$ True for all E $$2-E< \sqrt{4+\frac{1}{n}}$$ True for all n. So then the original compound inequality reduces to: $$4+\frac{1}{n}<(2+E)^{2}$$ $$4+\frac{1}{n}<4+4E+E^{2}$$ $$\frac{1}{n}<4E+E^{2}$$ $$n>\frac{1}{4E+E^{2}}$$ I am unsure as to the validity of the use of the transitive property above. – 2017-02-27

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