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).
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...
(1)
From (1) we derive...
(2)
... where...
(3)
... is analytic in and is . Consequence of that is that is also analytic and is...
(4)
... and finaly that is...
(5)
If we observe the (5) it is easy tom conclude that has a single pole is and here the residue is ...
Kind regards