# $\sum u_v$ converges absolutely iff $\sum \log(1 + u_v)$ converges absolutely

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?

• Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. – José Carlos Santos Apr 28 at 11:32
• You really don't need these inequalities to prove the statement in the title. All you need is the fact that $\frac {\log(1+z)} z \to 1$ as $z \to 0$ and this follows from the power series for $\log(1+z)$ (or by L'Hopital's Rule). – Kavi Rama Murthy Apr 28 at 11:40

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}
• Shouldn't it be $\log(1+u)$ in the numerator? – Botond Apr 28 at 12:37