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

\(\displaystyle \chi\) \(\displaystyle \sigma\)