How to prove this limit about $\gamma=\lim_{N\to \infty }\left(\sum_{n=1}^N\frac{1}{n}-\ln N\right)$ I have no idea how to prove it.
$$\lim_{m\rightarrow \infty }\left [ -\frac{1}{2m}+\ln\left ( \frac{e}{m} \right )+\sum_{n=2}^{m}\left ( \frac{1}{n}-\frac{\zeta \left ( 1-n \right )}{m^{n}} \right ) \right ]=\gamma $$
where $\gamma$ is the Euler Mascheroni constant and $\zeta$ is the Riemann zeta  function.
 A: One may recall the standard asymptotic expansion of the digamma function, as $X \to \infty$:
$$
\psi(X):=\frac{\Gamma'(X)}{\Gamma(X)}= \ln X - \frac{1}{2X} - \sum_{n=2}^m \frac{B_{n}}{n X^{n}}+\mathcal{O}\left(\frac1{X^{m+1}} \right), \quad m=2,3,\cdots. \tag1
$$ where $B_n$ are the Bernoulli numbers.
From $(1)$ and $\zeta(1-n)=-\dfrac{B_n}n,\, n=2,3,\cdots,$ $(2)$  and using $\displaystyle \sum_{n=1}^{m}\frac{1}{n}=\gamma+\psi(m+1)$ $(3)$, one obtains, as $m \to \infty$,
$$
\begin{align}
&-\frac{1}{2m}+\ln\left ( \frac{e}{m} \right )+\sum_{n=2}^{m}\left ( \frac{1}{n}-\frac{\zeta \left ( 1-n \right )}{m^{n}} \right )
\\\\&=\sum_{n=1}^{m}\frac{1}{n}-\ln m-\frac{1}{2m}+\sum_{n=2}^{m}\frac{B_{n}}{n}\frac{1}{m^n}
\\\\&=\gamma+\psi(m+1)-\ln m-\frac1m+\ln m-\psi(m)+\mathcal{O}\left(\frac1{m^{2}} \right)
\\\\&=\gamma+\mathcal{O}\left(\frac1{m^{2}} \right) \tag4
\end{align}
$$ where we have used $\displaystyle \psi(m+1)-\psi(m)=\frac1m,\, m=1,2,3,\cdots$. 
We obtain the announced limit.
A: This really just boils down to showing
$$\lim_{m\to\infty}\sum_{n=2}^m{\zeta(1-n)\over m^n}=0$$
since $-{1\over2m}$ clearly tends to $0$ and $\ln({e\over m})+\sum_{n=2}^m{1\over n}$ can be rewritten as $\sum_{n=1}^m{1\over n}-\ln m$, which tends to $\gamma$ more or less by definition of Euler's constant.
The key is the functional equation for the zeta function,
$$\zeta(1-n)={2\over(2\pi)^n}\Gamma(n)\zeta(n)\cos(\pi n/2)$$
from which it's easy, knowing that $\zeta(n)\lt2$ for $n\ge2$, to obtain the very crude inequality
$$|\zeta(1-n)|\le\Gamma(n)=(n-1)!\quad\text{for }n\ge2$$
We now see that
$$\begin{align}
\sum_{n=2}^m{|\zeta(1-n)|\over m^n}&\le{1!\over m^2}+{2!\over m^3}+{3!\over m^4}+\cdots+{(m-1)!\over m^m}\\\\
&={1\over m^2}\left(1+{2\over m}+{3\cdot2\over m\cdot m}+\cdots+{(m-1)(m-2)\cdots2\over m\cdot m\cdot m\cdots m} \right)\\\\
&\le{2\over m^2}(1+1+1+\cdots+1)\\\\
&={2(m-1)\over m^2}\\\\
&\le{2\over m}
\end{align}$$
and thus the limit tends to $0$ as $m\to\infty$.
