Behaviour of $\zeta(s)$ near $s=1$ I would appreciate if somebody could run this over and see if it works out? any suggestions or pointers would be appreciated. I denote the standard eta function $\eta$ by $\zeta^{*}$. I have not used big O notation and just used general well behaved functions. I do not wish to express the full error term, but instead ,just the principal part.
Behaviour of $\zeta(s)$ near $1$
From Abel's Theorem we can see that when $s=1$, $ \zeta^{*}(1) = \log(2)$. Now looking at $(1-2^{1-s})$ we can write it in terms of an exponential like so,
\begin{equation}
1-2^{1-s} = 1 - e^{(1-s)\log(2)}
\end{equation}
The power series expansion of $e^{z}$ is,
\begin{equation}
e^{z}= \sum_{n=0}^{\infty} \frac{z^{n}}{n!}\\
\Rightarrow  1-2^{1-s} = - e^{\log(2)(s-1)}=  - \sum_{n=0}^{\infty} \frac{((1-s)\log(2))^{n}}{n!}
\end{equation}
We can ignore the term when $n=0$ due to it being zero and sum from $n=1$ instead,
\begin{equation}
1-2^{1-s} = 0 - \sum_{n=1}^{\infty} \frac{(1-s)^{n}\log(2)^{n}}{n!}
\end{equation}
Expanding this sum and multiplying in the negative sign we have,
\begin{equation}
1-2^{1-s}= (s-1) \Bigg( \log(2) - \frac{\log(2)^{2}}{2!}(s-1) + \cdot \cdot \cdot \Bigg )
\end{equation}
Factorizing  the $\log(2)$ term out,
\begin{equation*}
(s-1)\log(2)\Bigg [ 1 - \bigg( \frac{\log(2)}{2!}(s-1) + \frac{\log(2)}{3!}(s-1)^{2} - \cdot \cdot \cdot  \bigg ) \Bigg ]
\end{equation*}
By the geometric series formula,  for $|s| < 1$, 
\begin{equation}
\frac{1}{\bigg[1 -  \bigg(  \frac{\log(2)}{2!}(s-1) + \cdot \cdot \cdot \bigg ) \bigg ] }= 1 + \Bigg( \frac{\log(2)}{2!}(s-1) + \cdot \cdot \cdot \Bigg ) +  \Bigg( \frac{\log(2)}{2!}(s-1) + \cdot \cdot \cdot \Bigg )^{2} +  \cdot \cdot \cdot 
\end{equation}
The terms of this geometric series decrease rapidly, so we are only  interested in keeping the first terms while letting a well-behaved and analytic function $g$ represent the remaining terms as a function in $s$.
\begin{equation}
\frac{1}{\bigg[1 -  \bigg(  \frac{\log(2)}{2!}(s-1) + \cdot \cdot \cdot \bigg ) \bigg ] } = 1 + \frac{\log(2)(s-1)}{2} + (s-1)^{2}\cdot g(s).
\end{equation}
We can now return to  $\frac{1}{1-2^{1-s}}$, and express  it in terms of what we have learned.
\begin{equation}
\frac{1}{1-2^{1-s}} = \frac{1}{\log(2)(s-1)} \Bigg(  1 + \frac{\log(2)(s-1)}{2} + (s-1)^{2}\cdot g(s) \Bigg ) 
= \frac{1}{\log(2)} \cdot \Bigg [ \frac{1}{s-1} + \frac{\log(2)}{2} + (s-1)g(s)\Bigg ]
\end{equation}
We can now study $\zeta(s)$ when $s$ is near to $1$. 
\begin{equation}
\zeta(s) = \frac{\zeta^{*}(s)}{1-2^{1-s}} =  \frac{\zeta^{*}(s)}{\log(2)} \cdot \Bigg [ \frac{1}{s-1} + \frac{\log(2)}{2} + (s-1)g(s)\Bigg ]
= \frac{\zeta^{*}(s)}{\log(2)} \cdot \frac{1}{s-1} + \frac{\zeta^{*}(s)}{2 \log(2)} \log(2) + \frac{\zeta^{*}(s)(s-1)g(s)}{\log(2)}
\end{equation}
As we know already, $\zeta^{*}(1) = \log(2)$ is analytic, so $\zeta^{*}(s)$ can be expanded as a series around $1$,
\begin{equation}
\zeta^{*}(s) = \log(2) + (s-1)a_1 + (s-1)^{2}a_2 +  \cdot \cdot \cdot
= \log(2) + (s-1) h(s)
\end{equation}
for a well behaved and analytic $h$. 
Near $s=1$ and by just looking at the principal terms,
\begin{equation}
\zeta(s) = \frac{\zeta^{*}(s)}{1-2^{1-s}} = \frac{ \log(2) +(s-1)h(s) }{\log(2)(s-1)}
= \frac{1}{s-1} + \frac{h(s)}{\log(2)}
\end{equation}

It looks fine to me. Maybe it's interesting to note that from $$\zeta\left(s\right)=s\int_{1}^{\infty}\frac{\frac{1}{2}-\left\{ x\right\} }{x^{s+1}}dx+\frac{1}{s-1}+\frac{1}{2},\,\,\textrm{Re}\left(s\right)>0$$ where $\left\{ x\right\}$ is the fractional part of $x$, we can easily get $$\zeta\left(s\right)=\frac{1}{s-1}+\gamma+O\left(\left|s-1\right|\right)$$ when $s$ is near to $1$.
