Answer
\begin{array}{lclcl}
b_1=1&\\
b_2=2&\\
b_3=\frac{5}{2}&\\
b_4=\frac{29}{10}&\\
\end{array}
Work Step by Step
Given $$ b_1=1, \ \ \ b_n=b_{n-1}+\frac{1}{b_{n-1}}, \ \ \ \ \ \ n= 2,3, \dots
$$
So, we get
\begin{array}{|l|l|}\hline
& n& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ b_n \\ \\ \hline
\text{First Term} & 1 & b_1=1 \\ \\ \hline
\text{Second Term} & 2 & b_2=b_{2-1}+\frac{1}{b_{2-1}}=b_{1}+\frac{1}{b_{1}}=1+\frac{1}{1}=2\\ \\ \hline
\text{Third Term} & 3 & b_3=b_{3-1}+\frac{1}{b_{3-1}}=b_{2}+\frac{1}{b_{2}}=2+\frac{1}{2}=\frac{5}{2}\\ \\ \hline
\text{Fourth Term} & 4 & b_4=b_{4-1}+\frac{1}{b_{4-1}}=b_{3}+\frac{1}{b_{3}}=\frac{5}{2}+\frac{1}{\frac{5}{2}}=\frac{5}{2}+\frac{2}{5}=\frac{29}{10}\\ \\ \hline
\end{array}
