How to prove this limit about $\gamma=\lim_{N\to \infty }\left(\sum_{n=1}^N\frac{1}{n}-\ln N\right)$

Question

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.

Answer 1

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.

Answer 2

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 $m\to\infty$.