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Let $$x_n = \frac{\sqrt{n^2+2}}{2n},n \in \mathbb{N^*} \ldots$$

Prove that $$|{x_n - \frac{1}{2}}| < \frac{1}{2n^2}$$

Indication: Use the relationship: $\sqrt{1+\gamma} < 1 + \frac{\gamma}{2}$, $ \forall \gamma \in \mathbb{R^{*}_{+}}$

I'd appreciate it very much if any answers could indicate a heuristic or general mindset one should have when proving such a proposition.

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3 Answers 3

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First of all we note that $\left|{\frac{\sqrt{n^2+2}}{2n} - \frac{1}{2}}\right| = {\frac{\sqrt{n^2+2}}{2n} - \frac{1}{2}}$ because ${\frac{\sqrt{n^2+2}}{2n} - \frac{1}{2}}>0$. In fact $\frac{\sqrt{n^2+2}}{2n}>\frac{1}{2} \Rightarrow \frac{n^2+2}{n^2}>1 \Rightarrow 1+\frac{2}{n^2}>1$.

Now we note that $\frac{\sqrt{n^2+2}}{2n} - \frac{1}{2}=\frac{\sqrt{\frac{2}{n^2}+1}}{2} - \frac{1}{2}$, and from indication $\sqrt{1+\gamma} < 1 + \frac{\gamma}{2}$, we have: $\frac{\sqrt{\frac{2}{n^2}+1}}{2} - \frac{1}{2}<\frac{\frac{1}{n^2}+1}{2} - \frac{1}{2}=\frac{1}{2n^2}$

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Observe that $(\sqrt{n^2+2}-n)(\sqrt{n^2+2}+n)=2$ and $\sqrt{n^2+2}+n>2n$, hence $0<\sqrt{n^2+2}-n<\frac 1{n} $ and finally $$0<\frac{\sqrt{n^2+2}}{2n}-\frac12<\frac 1{2n^2} . $$

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Hint: Bring the expression on the left to the common denominator $2n$. We get $$\left|\frac{\sqrt{n^2+2}-n}{2n}\right|.$$ Now multiply top and bottom by $\sqrt{n^2+2}+n$. The top simplifies, a lot.

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