I have two questions!
I request you to give me detailed solutions for the two!!!

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If you want to learn how to solve them or you are stuck at solving them, that's fine, we'll help you. We'll help you in understanding it or get past whatever you are stuck on.
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3. Hello, lutesium!

$\displaystyle \text{Evaluate: }\;S \;=\;\sum^n_{k=1}\ln\left(1 + \frac{1}{k}\right)$

$\displaystyle S \;=\;\ln\left(1 + \frac{1}{1}\right) + \ln\left(1 + \frac{1}{2}\right) + \ln\left(1 + \frac{1}{3}\right) + \ln\left(1 + \frac{1}{4}\right) + \hdots + \ln\left(1 + \frac{1}{n}\right)$

. . $\displaystyle =\;\ln(2) + \ln\left(\frac{3}{2}\right) + \ln\left(\frac{4}{3}\right) + \ln\left(\frac{5}{4}\right) + \hdots + \ln\left(\frac{n+1}{n}\right)$

. . $\displaystyle =\;\ln\left(2\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot \frac{5}{4}\:\cdots\;\frac{n+1}{n}\right)$

. . $\displaystyle =\;\ln\left(\frac{\rlap{/}2}{1}\cdot\frac{\rlap{/}3}{\rlap{/}2}\cdot\frac{\rlap{/}4}{\rlap{/}3}\cdot\frac{\rlap{/}5}{\rlap{/}4}\:\cdots\;\frac{n+1}{\rlap{/}n}\right)$

. . $=\;\ln(n+1)$

4. $e^{-x} + 1 = 0$

$e^{-x} = -1$

$-x = \log{(-1)}$

$-x = \ln{|-1|} + i\arg{(-1)}$

$-x = \ln{1} + i\pi$

$-x = 0 + i\pi$

$-x = i\pi$

$x = -i\pi$.