Firstly, we add $C $and $iS$ together $$
\begin{aligned}
C+iS&=\sum_{n=1}^{\infty}\left(\frac{e^i}{i}\right)^n \frac{1}{n}=-\ln \left(1-\frac{e^i}{i}\right) =-\ln \left(1+i e^i\right) \cdots(1)
\end{aligned}
$$
Secondly, we subtract $iS$ from $C$
$$
\begin{aligned}
C- & i S=-\ln \left(1-\frac{e^{-i}}{i}\right)=-\ln \left(1+i e^{-i}\right) \cdots(2)
\end{aligned}
$$
$(1)-(2)$ yields
$$
S=\frac{1}{2 i} \ln \left(\frac{1+i e^{-i}}{1+i e^i}\right)
$$
$$|S|=
\frac{1}{2}\left|\ln \left(\frac{1+i e^{-i}}{1+i e^i}\right)\right|$$
We are now going to express $\frac{1+i e^{-i}}{1+i e^i}$ in polar form.
$$
\begin{aligned}
\frac{1+i e^{-i}}{1+i e^i}& =\frac{1+e^{\frac{\pi-2 }{2}i}}{1+e^{\frac{\pi+2}{2} i}}
= \frac{e^{\frac{\pi-2}{4} i}\left(e^{\frac{\pi-2}{4} i}+e^{-\frac{\pi-2}{4} i}\right)}{e^{\frac{\pi+2}{4} i}\left(e^{\frac{\pi+2}{4} i}+e^{-\frac{\pi+2}{4} i}\right)} = e^{-i} \frac{\cos \left(\frac{\pi-2}{4}\right)}{\cos \left(\frac{\pi+2}{4}\right)}
\end{aligned}
$$
Then $$
\ln \left(\frac{1+i e^{-i}}{1+i e^i}\right) = \ln \left[\frac{\cos \left(\frac{\pi-2}{4}\right)}{\cos \left(\frac{\pi+2}{4}\right)}\right]-i
$$
We can now conclude that
$$
|S |=\frac{1}{2} \sqrt{\ln ^2\left[\frac{\cos \left(\frac{\pi-2}{4}\right)}{\cos \left(\frac{\pi+2}{4}\right)}\right]+1}
$$