# Why does this quotient involving the series representation of the Riemann Zeta function tend to $2$?

Let $$\zeta(n)$$ denote the Riemann Zeta function defined for positive integers $$n$$ as usual by:

$$\zeta(n)=\sum_{m=1}^{\infty} \frac{1}{m^n}.$$

It is known that for $$n=2$$ and $$3$$ there exists a series representation of $$\zeta$$ of the form:

$$\zeta (n)=\xi _{{n}}\sum _{{m=1}}^{{\infty }}{\frac {(-1)^{{m-1}}}{m^{{n}}{\binom {2m}{m}}}},$$

where $$\xi_n$$ is an algebraic number. We know that $$\xi_2 = \frac{\pi^2}{12 \sinh^{-1}(\frac{1}{2})^2}$$ and $$\xi_3=2.5$$. No such $$\xi_n$$ is known for $$n>3$$.

It is then natural to examine the behaviour of the function $$f:\mathbb{N}\to\mathbb{R}$$ defined by:

$$f(n) = \frac{\zeta (n)}{\sum _{{m=1}}^{{\infty }}{\frac {(-1)^{{m-1}}}{m^{{n}}{\binom {2m}{m}}}}}.$$

Curiously, it looks like $$\lim_{n\to\infty} f(n) = 2$$. I have tabulated the first values of $$f$$ below.

$$\begin{array}{|c|c|} \hline n & f(n) \\ \hline 2 & 3.55178\ldots \\ \hline 3 & 2.5 \\ \hline 4 & 2.20815\ldots \\ \hline 5 & 2.09487\ldots \\ \hline 6 & 2.04507\ldots \\ \hline 7 & 2.02187\ldots \\ \hline 8 & 2.01074\ldots \\ \hline 9 & 2.00531\ldots \\ \hline \end{array}$$

Is there a good explanation for this?

• "Natural"...for whom? – DonAntonio May 30 '18 at 9:53
• @Joanpemo I would think it pretty natural to want to study the coefficients $\xi_n$. Certainly, if you just define the $\xi_n$ from that series representation, they could be anything and wouldn't necessarily be so interesting. Once they've been proven to be algebraic, however, we know there is probably something special about them, and they become interesting. – Arthur May 30 '18 at 9:58
• Are all $\xi_n$ algebraic or is this proved for $n=2,3$ only? Can you give a reference? How can I see that $\xi_2$ is algebraic? – gammatester May 30 '18 at 12:38
• @gammatester: $\xi_3$ is rational because $\displaystyle \int_0^{1/2}\frac 4t\,\operatorname{arcsinh}(t)^2\,dt=\frac{2\,\zeta(3)}5\;$ (consider $x=\dfrac i2$ in $(4)$ here). I wouldn't bet on other algebraic solutions... An expression is provided too for $\xi_4$ but involves the "generalized log-sine integrals" $\operatorname{Ls}_4^{(1)}\,$ as well as Clausen functions. – Raymond Manzoni Oct 11 '18 at 22:43
• One of the references is Nan-Yue and Williams' "Values of the Riemann zeta function and integrals involving $\log\left(2\sinh\frac{\theta}2\right)$ and $\log\left(2\sin\frac{\theta}2\right)$" (see $(1.9)$). – Raymond Manzoni Oct 11 '18 at 22:44

## 1 Answer

The limit can be passed inside the summations, and the terms with $m\geq 2$ go to $0$. $$\lim_{n\to\infty}\zeta(n)=\sum_{m=1}^\infty \lim_{n\to\infty} \frac{1}{m^n}=1,$$ $$\lim_{n\to\infty}\sum _{{m=1}}^{{\infty }}{\frac {(-1)^{{m-1}}}{m^{{n}}{\binom {2m}{m}}}}=\sum _{{m=1}}^{{\infty }}\lim_{n\to\infty}{\frac {(-1)^{{m-1}}}{m^{{n}}{\binom {2m}{m}}}}=\frac{(-1)^0}{1{2\choose 1}}=\frac{1}{2}.$$