2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Nice solution! Thank you Olivier!₍₍⁽⁽(ી( ・◡・ )ʃ)₎₎⁾⁾ $\endgroup$ – Renascence_5. Feb 22 '16 at 0:16
1
$\begingroup$

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 $m\to\infty$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.