College Algebra, Chapter 9, 9.1, Section 9.1, Problem 38

Determine the first four partial sums and the $nth$ partial sum of the sequence $\displaystyle a_n = \frac{1}{n+1} - \frac{1}{n+2}$
The terms of the sequence are

$
\begin{equation}
\begin{aligned}
a_1 &= \frac{1}{1+1} - \frac{1}{1+2} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\\
\\
a_2 &= \frac{1}{2+1} - \frac{1}{2+2} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}\\
\\
a_3 &= \frac{1}{3+1} - \frac{1}{3+2} = \frac{1}{4} - \frac{1}{5} = \frac{1}{20}\\
\\
a_4 &= \frac{1}{4+1} - \frac{1}{4+2} = \frac{1}{5} - \frac{1}{6} = \frac{1}{30}\\
\end{aligned}
\end{equation}
$

The first four partial sums are.

$
\begin{equation}
\begin{aligned}
s_1 &= \frac{1}{6} = \frac{1}{2} - \frac{1}{3} &&= \frac{1}{6}\\
\\
s_2 &= \frac{1}{6} + \frac{1}{12} = \frac{1}{2} - \frac{1}{3} + \left(\frac{1}{3} - \frac{1}{4} \right) &&= \frac{1}{4}\\
\\
s_3 &= \frac{1}{6} + \frac{1}{12} + \frac{1}{20} = \frac{1}{2} - \frac{1}{3} + \left(\frac{1}{3} - \frac{1}{4} \right) + \left(\frac{1}{4} - \frac{1}{5} \right) &&= \frac{3}{10}\\
\\
s_4 &= \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} = \frac{1}{2} - \frac{1}{3} + \left(\frac{1}{3} - \frac{1}{4} \right) + \left(\frac{1}{4} - \frac{1}{5} \right) + \left(\frac{1}{5} - \frac{1}{6} \right) && = \frac{1}{3}
\end{aligned}
\end{equation}
$

The $nth$ partial sum is
$\displaystyle s_n = \frac{1}{2} - \frac{1}{n+2}$