Average employment length
This is simply the geometric distribution.

A job lasts 1 month with probability $s$.

A job lasts 2 months with probability $(1-s)s$.

A job lasts 3 months with probability $(1-s)^2s$.

$\vdots$

A job lasts $n$ months with probability $(1-s)^{n-1}s$.

Hence, the average length of a job is: $$\begin{align*}
 &amp; 1\cdot s+2\left(1-s\right)s+3\left(1-s\right)^{2}s+\dots+n\left(1-s\right){}^{n-1}s+\dots\\
= &amp; \left[s+\left(1-s\right)s+\left(1-s\right)^{2}s+\dots\right]+\left[\left(1-s\right)s+\left(1-s\right)^{2}s+\dots\right]+\dots\\
= &amp; \frac{s}{1-\left(1-s\right)}+\frac{1-s}{1-\left(1-s\right)}+\frac{\left(1-s\right)^{2}}{1-\left(1-s\right)}+\dots=\frac{s}{s}+\frac{1-s}{s}+\frac{\left(1-s\right)^{2}}{s}+\dots\\
= &amp; \frac{1}{s}\left(s+s\left(1-s\right)+s\left(1-s\right)^{2}+\dots\right)=\frac{1}{s}\frac{s}{1-\left(1-s\right)}=\frac{1}{s}.
\end{align*}$$

In the specific numerical example given by Mankiw, we have $s=0.01$ and hence the average length of a job is $1/s=100$.