# Help showing bound for magnitude of complex log fcn

1. Feb 25, 2006

### benorin

I'm working through an example problem wherein this bound is used:

$$\left| \log \left( 1-\frac{1}{L^s}\right) \right| \leq L^{-\sigma},$$

where $$s:=\sigma +it$$ and it is known that $$\sigma >1.$$ How do I prove this? Should I assume the principle brach is taken?

2. Feb 25, 2006

### benorin

It is perhaps important to note that in the context, $$L\rightarrow\infty$$

I think I got a start:

For $$\left| z\right| <1,$$ the Taylor series about z=0 is

$$\log (1-z)=-\sum_{j=1}^{\infty}\frac{z^j}{j}$$

so that

$$\left| \log \left( 1-\frac{1}{L^s}\right) \right| =\left| -\sum_{j=1}^{\infty}\frac{L^{-js}}{j} \right| \leq \sum_{j=1}^{\infty}\left| \frac{L^{-js}}{j}\right| = \sum_{j=1}^{\infty} \frac{L^{-j\sigma}}{j}$$

Last edited: Feb 25, 2006