Answer
\begin{array}{lclcl}
c_1=1&\\
c_2=\frac{3}{2}&\\
c_3=\frac{11}{6}&\\
c_4=\frac{25}{12}\\
\end{array}
Work Step by Step
Given $$\ \ \ c_n=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}, \ \ \ \ \ \ n= 1, 2,3, \dots
$$
So, we get
\begin{array}{|l|l|}\hline
& n& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ c_n \\ \\ \hline
\text{First Term} & 1 & c_1=1 \\ \\ \hline
\text{Second Term} & 2 & c_2=1+\frac{1}{2}=\frac{3}{2}\\ \\ \hline
\text{Third Term} & 3 & c_3=1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}\\ \\ \hline
\text{Fourth Term} & 4 & c_4=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{25}{12}\\ \\ \hline
\end{array}
