$\zeta(n+1) = \frac12 \Bigl(1+\zeta(n)\Bigr)$ for all integers $n\ge 28$

Question

It would appear that $$\zeta(n) = \frac12 \Bigl(1+\zeta(n-1)\Bigr)$$ holds true for all integers $n\ge 28$. I first seen this recurrence relation in Abramowitz and Stegun's Handbook of Mathematical Functions where there is a table of decimal approximations of $\zeta(n)$ from $1$ to $42$ and after that it says "For $n > 42, \zeta(n+1) = \frac12 \Bigl(1+\zeta(n)\Bigr)$ (page 811). However, after checking numerous other values in Wolfram Alpha, it appears this recurrence holds true as far back as $n=28$

However, searching the internet I cannot find anywhere that validates this statement. I would appreciate any sort of info on where I can learn more about this or if a proof can be produced. I for one do not know too much in depth about $\zeta(z)$ so I do not attempt to try proving on my own, although I would imagine Mathematical Induction is involved

Answer 1

Note that this is just an approximation to the zeta function using the first two terms in the series \begin{eqnarray*} \zeta(n) \sim 1+ \frac{1}{2^n}+ \cdots. \end{eqnarray*} which will give a good approximation because $1/3^n $ and higher terms will be very small for large values of $n$.

Answer 2

The formula listed is just an approximation. If it held exactly, we'd have \begin{align*} \zeta(2n + 2) = \frac{3}{4} + \frac{1}{4}\zeta(2n) \end{align*} for sufficiently large integers $n > 0$, but $\zeta(2m)$ is rational multiple of $\pi^{2m}$ for positive integers $m$, and $\pi$ is transcendental.