1
$\begingroup$

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

enter image description here

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?

$\endgroup$
  • $\begingroup$ Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. $\endgroup$ – José Carlos Santos Apr 28 at 11:32
  • 1
    $\begingroup$ 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). $\endgroup$ – Kavi Rama Murthy Apr 28 at 11:40
2
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ Thanks a lot! This was what I was looking for. Also thanks for the hint of not using pictures. $\endgroup$ – EinStone Apr 28 at 12:00
  • $\begingroup$ Shouldn't it be $\log(1+u)$ in the numerator? $\endgroup$ – Botond Apr 28 at 12:37
  • $\begingroup$ Of course! I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Apr 28 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.