# A formula for the Riemann zeta function at all positive integers

The function $f$ below claims to approximate the Riemann zeta function $\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}$ at all positive integers $s>1$:

$$f(s) = \frac{\pi^s}{\left\lfloor((2^s - 1)\frac{\pi^s}{2^s}\right\rfloor-1} \approx\zeta(s),$$

where $\lfloor x \rfloor$ represents the greatest integer small than or equal to $x$.

Indeed, tabulating the first few values gives:

\begin{array}{|c|c|c|} \hline s & f(s) & |\zeta(s) - f(s)| \\ \hline 2 & \frac{\pi^2}{6} & 0 \\ \hline 3 & \frac{\pi^3}{26} & 0.00950\ldots \\ \hline 4 & \frac{\pi^4}{90} & 0 \\ \hline 5 & \frac{\pi^5}{295} & 0.00042\ldots \\ \hline 6 & \frac{\pi^6}{945} & 0 \\ \hline 7 & \frac{\pi^7}{2995} & 0.00009\ldots \\ \hline 8 & \frac{\pi^8}{9450} & 0 \\ \hline 9 & \frac{\pi^9}{29749} & 0.000011\ldots \\ \hline \end{array}

I thought there was no function that even came close to "unifying" the values of $\zeta(s)$ at odd and even integers. What is the idea behind the construction of $f$ that allows such behaviour?

• What about $\zeta(12)$? – Angina Seng Apr 3 '18 at 14:33
• Did you calculate the decimal values of the give expressions for f(s), the second column, and compare them to the third column? They are nowhere near! For example, when s= 9, the middle column has $\frac{\pi^9}{29749}= 1.0020202$. That's not at all close to the third column, 0.000011. – user247327 Apr 3 '18 at 14:35
• @user247327 The third column is the difference between $\zeta$ and $f$... – Klangen Apr 3 '18 at 14:38
• @LordSharktheUnknown It is $2.302\ldots \times 10^{-7}$ – Klangen Apr 3 '18 at 14:39
• "I thought there was no function" It's an approximation, it's not a closed form. You can always create an approximation (and it's not hard to create an even better approximation). For example the simple expression $\zeta(s) \approx 1 + \frac{1}{2^s}$ is also a very good approximation for large $s$ (say all $s>10$). – Winther May 25 '18 at 12:50

For large $s$, $\zeta(s) \approx 1/(1 - 2^{-s})$ (this is the first factor of $\zeta$'s Euler product).
Let us approximate this approximation with something in the form $\pi^s/n(s)$ for some integer $n(s)$.

Then, $n(s)$ has to be close to $\pi^s(1-2^{-s})$. Naturally, picking $n(s) = \lfloor \pi^s(1-2^{-s})\rfloor$, we almost obtain your formula. In fact, if $\epsilon(s) = \pi^s(1-2^{-s}) - n(s)$, then

$\pi^s/n(s) = \pi^s/(\pi^s(1-2^{-s})-\epsilon(s)) = 1/(1 - 2^{-s} - \epsilon(s) \pi^{-s})\\ = 1 + 2^{-s} + \epsilon(s) \pi^{-s} + 4^{-s} + \ldots$

Substracting an extra $1$ from the denominator is the same as increasing $\epsilon(s)$ by $1$, and this makes the term $\epsilon(s) \pi^{-s}$ change into $(1+\epsilon(s))\pi^{-s}$, which I suppose is a bit closer to $3^{-s}$on average (especially for small values of $s$).

In both cases, the error term is of the order of $3^{-s}$ as $s$ gets large.

You can make better formulas (better because they work better for even values of $s$) by using more terms of the Euler product to build $n(s)$.

• Excellent answer, thank you. I can award you the bounty in "19 hours" :) – Klangen May 25 '18 at 13:01