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?
 A: 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 $n(s)$.
