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## Show the limit is zero

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 $\lim \sqrt{\dfrac{2}{n-1}}=0$.

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 $\lim \sqrt{\dfrac{2}{n-1}}=0$.