<s>[INST] Fractions in convergence proof Sequence: $a_n = \sqrt{2+ \frac{3}{n}}$
To prove convergence, want to show that $\left|\sqrt{2+ \frac{3}{n}} - \sqrt{2}\right| \le \varepsilon$
Simplifying, we get that $\sqrt{2+ \frac{3}{n}} - \sqrt{2}= \frac{3/n}{\sqrt{2 + \frac{3}{n}} + \sqrt{2}}$.
I understand up to this part. But then, the textbook says we need to replace this fraction by a larger fraction, for example show that $\frac{3/n}{2 \sqrt{2}} \le \varepsilon$
Why must we replace it by a larger fraction? Why is it not sufficient to say that the simplified fraction is less than epsilon? Can we choose any larger fraction that we want?
 [/INST] $\sqrt{2 + \frac{3}{n}}$ is larger than $\sqrt{2}$; then $\sqrt{2 +\frac{3}{n}} + \sqrt{2}$ is larger than $2\sqrt{2}$.  
So, $\frac{1}{\sqrt{2+\frac{3}{n}}+\sqrt{2}}$ is smaller than $\frac{1}{2\sqrt{2}}$ . So, now, you have an upper bound expressed in a simpler manner.  
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