Why does $\gamma=\lim_{s\to1^+}\sum_{n=1}^{\infty}\left(\frac{1}{n^s}-\frac{1}{s^n}\right)=\lim_{s\to0}\frac{\zeta(1+s)+\zeta(1-s)}{2}$? To be clear, I'm having trouble with proving both equalities, and would appreciate a hint. I'm also not sure why $1^+$ must be used as opposed to $1^-$. I'm not sure about the definition of $\zeta(x), x\le1$ (I encountered these equations here, to provide context). 
The first one reduces thusly
$$\gamma=\lim_{s\rightarrow 1^+}\sum_{n=1}^{\infty}\left(\frac{1}{n^s}-\frac{1}{s^n}\right)$$
$$=\lim_{s\rightarrow1^+}\left(\zeta(s)-\frac{\frac{1}{s}}{1-\frac{1}{s}}\right)=\lim_{s\rightarrow1^+}\left(\zeta(s)-\frac{1}{s-1}\right)$$
As $\gamma=\lim_{n\rightarrow \infty}\bigl(H_n-\ln(n)\bigr)$, the above equality is equivalent to
$$\lim_{s\rightarrow 0^+}\frac{1}{s}-\lim_{n\rightarrow \infty}\ln(n)=0$$
, although I am implicitly using
$$\lim_{n \rightarrow \infty}\sum_{k=1}^{n}\frac{1}{k}=\lim_{m \rightarrow 1^+}\sum_{k=1}^{\infty}\frac{1}{k^m}$$
which is may be wrong as the limits are approached differently. Regardless, I'm not sure how to progress from there.
I have even less of an idea about how to go about solving the second equality, perhaps because I have not dealt with antisymmetric limits before.
 A: As @David Speyer has proven the first part (see link in the comment), this answer focuses on the second part.
As you note, the problem is equivalent to proving:
$$\lim_{r\rightarrow1^+}\left(\zeta(r)-\frac{1}{r-1}\right)=\lim_{s\rightarrow 0^+}\left(\frac{\zeta(1+s)+\zeta(1-s)}{2}\right)$$
or equivalently:
$$\lim_{s\rightarrow 0^+}\left(\zeta(s+1)-\frac{1}{s}\right)=\lim_{s\rightarrow 0^+}\left(\frac{\zeta(1+s)+\zeta(1-s)}{2}\right) \tag{1}$$
By the  reflection formula for the Riemann zeta functionwe have:
$$\frac{\zeta  \left( s+1 \right)}{2} -{\pi }^{s}\cos \left( 1/2\,\pi \,s
 \right) \Gamma  \left( -s \right) \zeta  \left( -s \right) {2}^{s}=0
\tag{2}$$
and so subtracting $(2)$ from $(1)$ we obtain:
$$\lim_{s\rightarrow 0^+}\left(\frac{\zeta  \left( s+1 \right)}{2}-\frac{1}{s}+{\pi }^{s}\cos \left( 1/2\,
\pi \,s \right) \Gamma  \left( -s \right) \zeta  \left( -s \right) {2}
^{s}\right)=\lim_{s\rightarrow 0^+}\left(\frac{\zeta(1+s)+\zeta(1-s)}{2}\right)$$
and so it remains to prove:
$$\lim_{s\rightarrow 0^+}\left(\frac{\zeta  \left( 1-s \right)}{2} \right)=\lim_{s\rightarrow 0^+}\left(-\frac{1}{s}+{\pi }^{s}\cos \left( 1/2\,
\pi \,s \right) \Gamma  \left( -s \right) \zeta  \left( -s \right) {2}
^{s}\right) \tag{3}$$
Multiplying $(3)$ by $s$ we obtain:
$$\lim_{s\rightarrow 0^+}\left(\frac{s\zeta  \left( 1-s \right)}{2}\right)=\lim_{s\rightarrow 0^+}\left(-1+s{\pi }^{s}\cos \left( 1/2\,
\pi \,s \right) \Gamma  \left( -s \right) \zeta  \left( -s \right) {2}
^{s}\right) \tag{4}$$
and $(4)$ holds iff $(3)$ holds. Using the reflection formula again, $(4)$ becomes:
$$\lim_{s\rightarrow 0^+}\left(s\cos \left( 1/2\,\pi \,s \right) 
\Gamma  \left( s \right) \zeta  \left( s \right) {\pi }^{-s}{2}^{-s}
\right)=\lim_{s\rightarrow 0^+}\left(-1+s{\pi }^{s}\cos \left( 1/2\,
\pi \,s \right) \Gamma  \left( -s \right) \zeta  \left( -s \right) {2}
^{s}\right) \tag{5}$$
and we then note that:
$$s\Gamma(s)=\Gamma(s+1),\,\,\,-s\Gamma( -s)=\Gamma(-s+1)$$
both of which $\rightarrow 1$ as $s\rightarrow 0$ (as do the other trivial parts of $(5)$), and thus we are left to prove:
$$\lim_{s\rightarrow 0^+}\zeta(s)=\lim_{s\rightarrow 0^+}\left(-1-\zeta(-s)\right) \tag{6}$$
The well known limit:
$$\lim_{s\rightarrow 0^+}\zeta(s)=-\frac{1}{2}$$
is proven here and the limit is the same from both directions as the Riemann Zeta function is meromorphic; the only pole being at $\zeta(1)$, and thus both sides of $(6)$ are equal to $-1/2$ which proves $(5),(3)$ and hence $(1)$.
A: Dave Renfro's paper ’Euler's Constant $\gamma$' and David Speyer's link should be helpful for an elementary derivation of the first part i.e. get the limit :
$$\tag{0}\gamma=\lim_{s\rightarrow 1^+}\left[\zeta(s)-\frac{1}{s-1}\right]$$ 
As indicated by Gerry your problem was to go from a well defined limit (the limit of the difference $\,\zeta(s)-\frac{1}{s-1}\,$ as $\,s\rightarrow1^+$) to the difference of the limits when these limits are both infinite !
$$-$$
Concerning your second limit :
$$\tag{1}\gamma=\lim_{s\to0}\frac{\zeta(1+s)+\zeta(1-s)}{2}$$
this will require a better definition of $\zeta\,$ than $\,\displaystyle\zeta(s):=\sum_{n=1}^{\infty} \frac{1}{n^s}$ since this definition is valid only for $\Re(s)>1$.
To go further you may use the Dirichlet eta function with the idea of converting a sum of positive terms to an alternate sum so that $\zeta$ may then be written as :
$$\tag{2}\zeta(s)=-\frac 1{1-2^{1-s}}\sum_{n=1}^{\infty} \frac{(-1)^n}{n^s}$$
which is convergent for any complex $s$ such that $\,\Re(s)>0,\ s\not =1\,$ or better use the analytic extension of $\zeta$ in the whole complex plane except $s=1$ where $\zeta\,$ has a simple pole (as you found).  
To see how to obtain the alternate series $(2)$ (and convergence proof) as well as get some intuitive ideas about analytic continuation of $\zeta\,$ you may see this answer.
Let's note that once the Laurent series of $\zeta$ at $s=1$ obtained with the simple pole at $1$ :
$$\tag{3}\zeta(s)=\frac 1{s-1}+\sum_{n=0}^\infty \frac{(-1)^n}{n!}\gamma_n\;(s-1)^n$$
with $\gamma_n$ the Stieltjes constants and $\gamma_0=\gamma$ your Euler constant then the limit of $\zeta(s)-\frac 1{s-1}$ at $s\to 1$ is rather straightforward.
Using the alternate series or the analytic extension you'll get that the limit was in fact given by (note that $\,s\rightarrow1^+$ was replaced by $\,s\rightarrow1$) :
$$\gamma=\lim_{s\rightarrow 1}\left[\zeta(s)-\frac{1}{s-1}\right]=\lim_{z\rightarrow 0}\left[\zeta(1+z)-\frac{1}z\right]$$
The idea is simply to rewrite the $z$ at the right as $+s$ and as $-s$ and to return the mean value to get :
$$\gamma=\lim_{s\rightarrow 0}\frac 12\left[\left(\zeta(1+s)-\frac{1}s\right)+\left(\zeta(1-s)-\frac{1}{-s}\right)\right]$$
or
$$\gamma=\lim_{s\rightarrow 0}\frac {\zeta(1+s)+\zeta(1-s)}2$$
Let's conclude with an elementary proof using the alternate series $(2)$ :
\begin{align}
\zeta(1+s)+\zeta(1-s)&=\sum_{n=1}^{\infty}\frac 1{2^{-s}-1} \frac{(-1)^n}{n^{1+s}}+\frac 1{2^{s}-1} \frac{(-1)^n}{n^{1-s}}\\
&=\sum_{n=1}^{\infty}\frac{(-1)^n}n\left[\frac 1{2^{-s}-1} \frac 1{n^{s}}+\frac 1{2^{s}-1} \frac 1{n^{-s}}\right]\\
&=\sum_{n=1}^{\infty}\frac{(-1)^n}n\left[\left(e^{-s\ln(2)}-1\right)^{-1}e^{-s\ln(n)} +\left(e^{s\ln(2)}-1\right)^{-1}e^{s\ln(n)}\right]\\
&=\sum_{n=1}^{\infty}\frac{(-1)^n}n\left[2\frac{\ln(n)}{\ln(2)}-1+\sum_{m=1}^\infty s^{2m}\,P_{2m}(\ln(m))\right]\\
\end{align}
with $P_{2m}$ polynomials depending of $\ln(n)$ and constants only.
But $\lim_{n\to\infty}\frac{\ln(n)^k}n=0$ for any nonnegative integer $k$ so that we get another nice series equal to $\gamma$ : 
$$
\lim_{s\rightarrow 0}\frac {\zeta(1+s)+\zeta(1-s)}2=\sum_{n=1}^{\infty}\frac{(-1)^n}n\,\left(\frac{\ln(n)}{\ln(2)}-\frac 12\right)=\gamma$$
