Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 9 Exercise 9.7

Solution: We have seen that $0\le s_n<\sqrt{\dfrac{2}{n-1}}$. By Sequeeze-Theorem/Exercise 8.5, it suffices to show that $\lim \sqrt{\dfrac{2}{n-1}}=0$.

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Let $\epsilon>0$. Let $N>\dfrac{2}{\epsilon^2}+1$. If $n>N$, we have

$$

n-1>\frac{2}{\epsilon^2}\Longrightarrow \sqrt{\dfrac{2}{n-1}}<\epsilon.

$$ Therefore, we have

$$

\left|\sqrt{\dfrac{2}{n-1}}-0\right|<\epsilon

$$ for all $n>N$ as desired. Thus $\lim \sqrt{\dfrac{2}{n-1}}=0$ and we are done.

A different approach to see that it suffices to show that $\lim \sqrt{\dfrac{2}{n-1}}=0$ is as follows.

We know that if $\lim s_n$ exists, by Exercise 8.9, we must have

$$

0\le \lim s_n\le \lim \sqrt{\dfrac{2}{n-1}}.

$$ Hence it suffices to show that $\lim \sqrt{\dfrac{2}{n-1}}=0$.

$$

0\le \lim s_n\le \lim \sqrt{\dfrac{2}{n-1}}.

$$ Hence it suffices to show that $\lim \sqrt{\dfrac{2}{n-1}}=0$.