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$\displaystyle \displaystyle \text{Evaluate: }\;S \;=\;\sum^n_{k=1}\ln\left(1 + \frac{1}{k}\right)$
$\displaystyle \displaystyle S \;=\;\ln\left(1 + \frac{1}{1}\right) + \ln\left(1 + \frac{1}{2}\right) + \ln\left(1 + \frac{1}{3}\right) + \ln\left(1 + \frac{1}{4}\right) + \hdots + \ln\left(1 + \frac{1}{n}\right) $
. . $\displaystyle \displaystyle =\;\ln(2) + \ln\left(\frac{3}{2}\right) + \ln\left(\frac{4}{3}\right) + \ln\left(\frac{5}{4}\right) + \hdots + \ln\left(\frac{n+1}{n}\right) $
. . $\displaystyle \displaystyle =\;\ln\left(2\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot \frac{5}{4}\:\cdots\;\frac{n+1}{n}\right) $
. . $\displaystyle \displaystyle =\;\ln\left(\frac{\rlap{/}2}{1}\cdot\frac{\rlap{/}3}{\rlap{/}2}\cdot\frac{\rlap{/}4}{\rlap{/}3}\cdot\frac{\rlap{/}5}{\rlap{/}4}\:\cdots\;\frac{n+1}{\rlap{/}n}\right) $
. . $\displaystyle =\;\ln(n+1)$
$\displaystyle e^{-x} + 1 = 0$
$\displaystyle e^{-x} = -1$
$\displaystyle -x = \log{(-1)}$
$\displaystyle -x = \ln{|-1|} + i\arg{(-1)}$
$\displaystyle -x = \ln{1} + i\pi$
$\displaystyle -x = 0 + i\pi$
$\displaystyle -x = i\pi$
$\displaystyle x = -i\pi$.