I had originally found that $\lfloor\zeta(\zeta(n))\rfloor$, where $\zeta(n)$ is the Riemann Zeta Function, seemed to be relatively close to $\left\lfloor\frac{1}{\zeta(n)-1}\right\rceil$ for $n \in \mathbb{Z} $ such that $n \geq 2$. This led me to realize that for real values of $x$ between $1$ and $2$, $\zeta(n) \approx \frac{1}{x-1}$. I posted a question about this here. Thanks to Nilotpal Kanti Sinha's response to that post, I was informed the similarities between the two expressions is because of the Laurent series expansion of the Riemann Zeta function:
$$\zeta(s) = \frac{1}{s-1} + \sum_{n=0}^\infty \frac{(-1)^n}{n!} \gamma_n (s-1)^n$$
where $\gamma_n$ is the $n$-th Stieltjes constant. I had also noticed that on the same interval, though, it seems that $\zeta(n)>\frac{1}{x-1}$. I tried proving this by first realizing this would just be the same as proving that, if $1<x<2$,
$$\sum_{n=0}^\infty \frac{(-1)^n}{n!} \gamma_n (x-1)^n > 0$$
I don't know how to continue, though, as I don't fully understand the Stieltjes constants. Is it possible to prove or disprove this? Is it possible to prove this inequality for all $x>1$? Additionally, can it be proven that the above infinite sum expression is strictly increasing on the interval $x \in (1,2)$?
The inequality is correct, even for all $x>0$ (except $x=1$ where $\zeta(x)$ is not defined).
We start with the representation $$ \zeta(s) = \frac s{s-1} - s\int_1^\infty \{u\} u^{-s-1}\,du $$ valid for $\Re s>0$; here $\{u\}=u-\lfloor u\rfloor$ is the fractional part. (Note that this representation immediately implies that $\zeta(x) - 1/(x-1)$ is strictly increasing for $x>0$.)
Using $0\le\{u\}<1$ in the integrand, and setting $s=x$ to be a positive real number, yields $$ 0 \le \int_1^\infty \{u\} u^{-x-1}\,du < \int_1^\infty u^{-x-1}\,du = \frac1x; $$ therefore $$ \zeta(x) = \frac x{x-1} - x\int_1^\infty \{u\} u^{-x-1}\,du > \frac x{x-1} - x \cdot\frac1x = \frac1{x-1} $$ and, for good measure, $$ \zeta(x) = \frac x{x-1} - x\int_1^\infty \{u\} u^{-x-1}\,du \le \frac x{x-1}, $$ both valid for all $x\in(0,\infty)\setminus\{1\}$.