A limit about euler's constant Show that :
$$\lim_{m\to \infty}\left[ -\frac{1}{2m}+\ln \left( \frac{\text{e}}{m} \right)+\sum\limits_{n=2}^m \left( \frac{1}{n}-\frac{\zeta \left( 1-n \right)}{m^n} \right) \right]=\gamma $$
How to evaluate that sum?
 A: This is an expression of the asymptotic series for $1+\frac12+\dots+\frac1m$,
$$
1+\frac12+\dots+\frac1m= \ln{m}+\gamma+\frac{1}{2m}-\sum_{k=1}^r \frac{B_{2k}}{2k m^{2k}}+{\cal R}(m, r),
$$
where $-B_{2k}/(2k)$ has been replaced by the equal value $\zeta(1-2k)$.  (Since $\zeta(-s)$ is zero for $s$ positive and even, these terms can be omitted.)  The term inside the limit is $$\gamma+{\cal R}(m, \lfloor m/2 \rfloor).$$
From [1], and using $1+\dots+\frac1m=\psi(m)+\gamma+\frac1m$,
$$
|{\cal R}(m,r)|\le \frac{|B_{2r+2}|}{(2r+2) m^{2r+2}} \qquad (1)
$$
for $m>0$ and $r\ge 0$.  From [2],
$$
|B_{2n}|\sim 4\sqrt{\pi n} (\frac{n}{\pi e})^{2n}, \qquad n\to\infty.\qquad (2)
$$
Combining (1) and (2) gives
$$
|{\cal R}(m,\lfloor m/2 \rfloor)|\le (1+o(1))
2\sqrt{\frac{2\pi}{m}} e^{2\lfloor m/2\rfloor+2-m} (2\pi e)^{-2\lfloor m/2 \rfloor-2}.\qquad (3)
$$
Since the right-hand side of (3) goes to $0$ as $m\to\infty$, this proves that the limit is $\gamma$, as desired.
A: To show that
$$
\lim_{m\to\infty}\left[-\frac1{2m}+\log\left(\frac em\right)+\sum_{n=2}^m \left(\frac1n-\frac{\zeta(1-n)}{m^n}\right)\right]=\gamma\tag{1}
$$
we pretty simply have
$$
\lim_{m\to\infty}\left[-\frac1{2m}\right]=0\tag{2}
$$
and
$$
\begin{align}
\lim_{m\to\infty}\left[\log\left(\frac em\right)+\sum_{n=2}^m\frac1n\right]
&=\lim_{m\to\infty}\left[\sum_{n=1}^m\frac1n-\log(m)\right]\\[12pt]
&=\gamma\tag{3}
\end{align}
$$
So all that remains is to show that
$$
\lim_{m\to\infty}\left[\sum_{n=2}^m\frac{\zeta(1-n)}{m^n}\right]=0\tag{4}
$$
Define
$$
a_m(n)=\left\{\begin{array}{}
4\sqrt{\pi/n}\left(\frac{n}{\pi me}\right)^{2n}&\text{when }n\le\lfloor m/2\rfloor\\
4\sqrt{\pi/n}\left(\frac1{2\pi e}\right)^{2n}&\text{when }n\gt\lfloor m/2\rfloor
\end{array}\right.\tag{5}
$$
Note that
$$
\sum_{n=1}^\infty a_0(n)\le\frac{4\sqrt\pi}{4\pi^2e^2-1}\tag{6}
$$
and $a_m(n)\to0$ monotonically as $m\to0$. Thus, by Dominated Convergence,
$$
\lim_{m\to\infty}\sum_{n=1}^\infty a_m(n)=0\tag{7}
$$
Since
$$
\begin{align}
\left|\,\sum_{n=2}^m\frac{\zeta(1-n)}{m^n}\,\right|
&=\left|\,\sum_{n=1}^{\lfloor m/2\rfloor}\frac{\zeta(1-2n)}{m^{2n}}\,\right|\\
&=\left|\,\sum_{n=1}^{\lfloor m/2\rfloor}-\frac{B_{2n}}{2n\,m^{2n}}\,\right|\\
&=\left|\,\sum_{n=1}^{\lfloor m/2\rfloor}(-1)^n\zeta(2n)\frac{(2n)!}{n(2\pi m)^{2n}}\,\right|\\
&\le\sum_{n=1}^{\lfloor m/2\rfloor}2\frac{\sqrt{4\pi n}(2n)^{2n}e^{-2n}}{n(2\pi m)^{2n}}\\
&=\sum_{n=1}^{\lfloor m/2\rfloor}4\sqrt{\pi/n}\left(\frac{n}{\pi me}\right)^{2n}\tag{8}
\end{align}
$$
we see that $(7)$ verifies $(4)$ and we are done.
