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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?

user127.0.0.1
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    You don't have to but it is easier to find an $n$ that fullfills the inequality this way – user127.0.0.1 Jan 16 '14 at 02:24
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    It's often easier to prove a looser upper bound on the difference, as long as that upper bound still tends to zero. Choosing the form of the upper bound is a bit of an art, but keeping it simple makes it easier to find an $N$ such that (the upper bound on) the difference is less than $\varepsilon$ for all $n\ge N$. – mjqxxxx Jan 16 '14 at 02:27
  • But why is it not correct to simplify $\frac{3/n}{\sqrt{2 + \frac{3}{n}} + \sqrt{2}} \le \epsilon$ and get the expression for n? It seems that in all proofs, n is chosen based on the larger fraction. –  Jan 16 '14 at 02:27
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    It is not incorrect but it's harder to find such an n, as already mentioned – user127.0.0.1 Jan 16 '14 at 02:29
  • So the only issue is that using the simplified fraction would be more difficult algebraically than using the approximation? –  Jan 16 '14 at 02:49

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$\sqrt{2 + \frac{3}{n}}$ 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.

janmarqz
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Claude Leibovici
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