How can I prove that
$\displaystyle \frac{1}{1-2^{1-s}}$
has a simple pole at s=1.
Any help is much appreciated.
Here is a clue: Read this limt of (x - 1)/(1 - 2^(1-x)) as x approaches 1 - Wolfram|Alpha (click on Show steps if you don't know how to get the limit).
Hi,
You could show the pole has order 1 directly:
Pole (complex analysis) - Wikipedia, the free encyclopedia
Perhaps another possible method is as follows, though I could be wrong:
I have just posted up a question regarding the domain of the Riemann zeta function and am aware that we can show it has a singularity at s = 1 by rewriting it as the product of an infinite sum by the term you have posted up. The analytic continuation of the Riemann zeta function, valid in Re(z) > 0 gives it as 1 + 1/(s-1) + {some integral well defined integral} and hence we see straight away that is has a simple pole at s = 1.
Lets start considering that is...
$\displaystyle 2^{1-s}= e^{(1-s) \ln 2} = \sum_{n=0}^{\infty} \frac{(-1)^{n} \ln^{n} 2}{n!} (s-1)^{n} $ (1)
From (1) we derive...
$\displaystyle 1-2^{1-s} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \ln^{n} 2}{n!} (s-1)^{n}=$
$\displaystyle = (s-1) \sum_{n=0}^{\infty} \frac{(-1)^{n} \ln^{n+1} 2}{(n+1)!} (s-1)^{n} = (s-1)\cdot \ln 2 \cdot \varphi(s)$ (2)
... where...
$\displaystyle \varphi(s) = 1 - \frac{\ln 2}{2} \cdot (s-1) + ... $ (3)
... is analytic in $\displaystyle s=1$ and is $\displaystyle \varphi(1)=1$. Consequence of that is that $\displaystyle \frac{1}{\varphi(s)}$ is also analytic and is...
$\displaystyle \frac{1}{\varphi(s)} = 1 + \frac{\ln 2}{2} \cdot (s-1) + ... $ (4)
... and finaly that is...
$\displaystyle \frac{1}{1-2^{1-s}}= \frac{1}{\ln 2 \cdot (s-1)\cdot \varphi(s)} = \frac{1}{\ln 2 \cdot (s-1)} \{ 1 + \frac{\ln 2}{2} \cdot (s-1) + ...\} $ (5)
If we observe the (5) it is easy tom conclude that $\displaystyle \frac{1}{1-2^{1-s}}$ has a single pole is $\displaystyle s=1$ and here the residue is $\displaystyle \frac{1}{\ln 2}$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$