I have trouble understanding the following proof of a fact in complex analysis.
Assume $(u_v)_{v\geq1}$ is a sequence of complex numbers and $(u_v) \neq -1$ for all $v$. Then we have the following
In particular I want to understand the estimate $\frac23|u| \leq\log(1+u)| \leq \frac43|u|$. I know that for $|u| < 1$ the power series expansion of $\log(1+u)$ is: $$\log(1+u) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} u^n$$
Triangle inequality doesn't seem to help, but how can I derive the estimates?
We have, if $u\neq 0$,\begin{align}\left\lvert\frac{\log(1+u)}u\right\rvert&=\left\lvert1-\frac u2+\frac{u^2}3-\frac{u^3}4+\cdots\right\rvert\\&\leqslant1+\lvert u\rvert+\lvert u\rvert^2+\lvert u\rvert^3+\cdots\\&\leqslant1+\frac14+\left(\frac14\right)^2+\left(\frac14\right)^3+\cdots\\&=\frac43\end{align}and, on the other hand:\begin{align}\left\lvert\frac{\log(1+u)}u\right\rvert&=\left\lvert1-\frac u2+\frac{u^2}3-\frac{u^3}4+\cdots\right\rvert\\&\geqslant1-\left\lvert\frac u2+\frac{u^2}3-\frac{u^3}4+\cdots\right\rvert\\&=1-\lvert u\rvert\left\lvert\frac 12+\frac u3-\frac{u^2}4+\cdots\right\rvert\\&\geqslant1-\lvert u\rvert\left(1+\frac14+\left(\frac14\right)^2+\left(\frac14\right)^3+\cdots\right)\\&\geqslant1-\frac14\times\frac43\\&=\frac23.\end{align}