1. Simple Poles

How can I prove that

$\frac{1}{1-2^{1-s}}$

has a simple pole at s=1.

Any help is much appreciated.

2. Originally Posted by signature
How can I prove that

$\frac{1}{1-2^{1-s}}$

has a simple pole at s=1.

Any help is much appreciated.
Here is a clue: Read this limt of &#x28;x - 1&#x29;&#x2f;&#x28;1 - 2&#x5e;&#x28;1-x&#x29;&#x29; as x approaches 1 - Wolfram|Alpha (click on Show steps if you don't know how to get the limit).

3. Simple Poles

I still dont understand it.

Is there no simple method to approach it,

4. Originally Posted by signature
I still dont understand it.

Is there no simple method to approach it,
Go to your classnotes. In them you will find a theorem that says that the singularity $z = z_0$ is a simple pole of f(z) if $\lim_{z \to z_0} ((z - z_0) f(z))$ exists and is finite. This is the simple approach I have suggested that you use.

5. 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.

6. Lets start considering that is...

$2^{1-s}= e^{(1-s) \ln 2} = \sum_{n=0}^{\infty} \frac{(-1)^{n} \ln^{n} 2}{n!} (s-1)^{n}$ (1)

From (1) we derive...

$1-2^{1-s} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \ln^{n} 2}{n!} (s-1)^{n}=$

$= (s-1) \sum_{n=0}^{\infty} \frac{(-1)^{n} \ln^{n+1} 2}{(n+1)!} (s-1)^{n} = (s-1)\cdot \ln 2 \cdot \varphi(s)$ (2)

... where...

$\varphi(s) = 1 - \frac{\ln 2}{2} \cdot (s-1) + ...$ (3)

... is analytic in $s=1$ and is $\varphi(1)=1$. Consequence of that is that $\frac{1}{\varphi(s)}$ is also analytic and is...

$\frac{1}{\varphi(s)} = 1 + \frac{\ln 2}{2} \cdot (s-1) + ...$ (4)

... and finaly that is...

$\frac{1}{1-2^{1-s}}= \frac{1}{\ln 2 \cdot (s-1)\cdot \varphi(s)} = \frac{1}{\ln 2 \cdot (s-1)} \{ 1 + \frac{\ln 2}{2} \cdot (s-1) + ...\}$ (5)

If we observe the (5) it is easy tom conclude that $\frac{1}{1-2^{1-s}}$ has a single pole is $s=1$ and here the residue is $\frac{1}{\ln 2}$...

Kind regards

$\chi$ $\sigma$