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Solution:

Part a

$s_1=1$, $s_2=\sqrt{2}$, $s_3=\sqrt{\sqrt 2+1}$, $s_4=\sqrt{\sqrt{\sqrt 2+1}+1}$.

Part b

Since $(s_n)$ converges. Set $\lim s_n=s$. By $s_{n+1}=\sqrt{s_n+1}$, we have
$$
s_{n+1}^2=s_n+1.
$$ Hence we have
\begin{equation}\label{09-04-01}
\lim s_{n+1}^2=\lim (s_n+1).
\end{equation} On one hand, by Theorem 9.4, we have
\begin{equation}\label{09-04-02}
\lim s_{n+1}^2=(\lim s_{n+1})^2=s^2.
\end{equation} On the other hand, by Theorem 9.3, we have
\begin{equation}\label{09-04-03}
\lim(s_n+1)=\lim s_n+\lim 1=s+1.
\end{equation} Combining \eqref{09-04-01}, \eqref{09-04-02}, \eqref{09-04-03}, we have
$$
s^2=s+1.
$$ Solving for $s$, we must have $s=\dfrac{1\pm \sqrt 5}{2}$.

Since $s_n>0$ for all $n$, it follows from Exercise 8.9 that $s>0$. Therefore, we conclude that $s=\dfrac{1+\sqrt 5}{2}$.