We Know ($ z \in \textbf{C} $):
$\dfrac{1}{1-z} = \sum_{n=0}^{k} z^{n} + \dfrac{z^{k+1}}{1-z}$
Integrating this along the straight line $L$ from $0$ to $z$:
$ -ln(1-z) = \sum_{n=0}^{k}\dfrac{z^{n+1}}{n+1} + \int_{L}\dfrac{w^{k+1}}{1-w} dw$
If $ \vert z \vert < 1$, an upper bound for the absolute value of the integral is:
$ \vert \int_{L} \dfrac{w^{k+1}}{1-w} dw \vert \leq \vert z \vert \dfrac{\vert z \vert^{k+1}}{1 - \vert z \vert} = \dfrac{\vert z \vert^{k+2}}{1-\vert z \vert}$
From here we get:
$\sum_{n=1}^{\infty}\dfrac{z^{n}}{n} = - ln(1-z)$, $ \quad \vert z \vert<1 $.
Can someone tell me how the upper bound was found?
Just looking at the real case:
If $0 < z < 1$, then there are two different upper bounds.
$\begin{array}\\ \left|\int_{0}^z\dfrac{w^{k+1}}{1-w} dw\right| &\le \left|\int_{0}^z\dfrac{w^{k+1}}{1-z} dw\right| \qquad z \ge w \implies 1-z \le 1-w\\ &=\dfrac1{1-z} \left|\int_{0}^zw^{k+1} dw\right|\\ &=\dfrac1{1-z} \left|\dfrac{z^{k+2}}{(k+2)}\right|\\ &=\dfrac{z^{k+2}}{(1-z)(k+2)}\\ \end{array} $
$\begin{array}\\ \left|\int_{0}^z\dfrac{w^{k+1}}{1-w} dw\right| &\le \left|\int_{0}^z\dfrac{z^{k+1}}{1-w} dw\right| \qquad z \ge w\\ &= z^{k+1}\int_{0}^z\dfrac{1}{1-w} dw\\ &= z^{k+1}\int_{1-z}^1\dfrac{1}{w} dw\\ &= z^{k+1}(\ln(w))|_{1-z}^1\\ &= z^{k+1}(-\ln(1-z))\\ \end{array} $
and then use whatever bounds you want based on $-\ln(1-z) =\sum_{k=1}^{\infty} \dfrac{z^k}{k} $.