This is the Sum:
$$\sum\limits_{n=1}^{\infty} \frac{\zeta(2n+1)-1}{\zeta(2n)-1}$$
Sum is divergent, but Wolfram acts weird and after a correct display of early values, it starts converging to a random value: Wolfram Alpha (Link computes the Sum up to $n=40$, just to get a faster result);
Probably just a bug.
There are obviously some numerical issues.
Consider $$S_p=\sum\limits_{n=1}^{p} \frac{\zeta(2n+1)-1}{\zeta(2n)-1}$$ Up to $p=23$, we have with a quite good accuracy $$S_p \sim \frac {2p-1}4$$ For $p=23$, the exact value is $11.2656$ while the approximation gives $11.25$. However for $p=24$, a CAS reports a sum equal to $13.2656$ while the approximation would give $11.75$.
Where the problem is that, numerically, computed separately with unlimited precision $$\zeta(49) \sim 1+1.776\times 10^{-15}$$ $$\zeta(48) \sim 1+3.553\times 10^{-15}$$ which would give
$$\frac{\zeta(49)-1}{\zeta(48)-1} \sim 0.49999999941187591$$ Computing the above as a whole leads to the same result. But, nevertheless, we are very close to machine $\epsilon$.
Computed seperately, the summand is asymptotic to $0.5^-$.
We can also use the double inequality (have a look at @reuns comment here)
$$1+2^{-k}+\frac{3^{1-k}}{k-1} < \zeta(k) < 1+2^{-k}+\frac{2^{1-k}}{k-1}$$ which gives $$2^{-2 n-1}+\frac{3^{-2 n}}{2 n}<\zeta(2n+1)-1 <2^{-2 n-1}+\frac{2^{-2 n}}{2n}$$ $$2^{-2 n}+\frac{3^{1-2 n}}{2 n-1}<\zeta(2n)-1<2^{-2 n}+\frac{2^{1-2 n}}{2 n-1}$$ from which $$\frac{2^{-2 n-1}+\frac{3^{-2 n}}{2 n}}{2^{-2 n}+\frac{2^{1-2 n}}{2 n-1}}<\frac{\zeta(2n+1)-1 }{\zeta(2n)-1 }< \frac{2^{-2 n-1}+\frac{2^{-2 n-1}}{n}}{2^{-2 n}+\frac{3^{1-2 n}}{2 n-1}}$$
Using $p=100$ this gives $47.8309 < S_{100} <51.8495 $. Notice that the mean of these bounds is $49.84$ while $\frac {199}4=49.75$.
Moreover, in the attached Wolfram Alpha, $S_{40}$ is given as $\sim 19.73$ which does not show at all in the plot ! The above would give $18.2817 < S_{40} <21.3951 $ (mean=$19.84$) while $\frac {79}4=19.75$