Simple Poles

Nov 2009
How can I prove that

\(\displaystyle \frac{1}{1-2^{1-s}}\)

has a simple pole at s=1.

Any help is much appreciated.
Nov 2009
I still dont understand it.

Is there no simple method to approach it,

mr fantastic

MHF Hall of Fame
Dec 2007
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 \(\displaystyle z = z_0\) is a simple pole of f(z) if \(\displaystyle \lim_{z \to z_0} ((z - z_0) f(z))\) exists and is finite. This is the simple approach I have suggested that you use.
  • Like
Reactions: signature
May 2010

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.
  • Like
Reactions: signature


MHF Hall of Honor
Mar 2009
near Piacenza (Italy)
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

\(\displaystyle \chi\) \(\displaystyle \sigma\)
  • Like
Reactions: signature