Desde $ 1 < \zeta(s) <2$ $s \ge 2$, es equivalente a indicar que $ \displaystyle\sum_{s=2}^{\infty} \left( \zeta(s)-1 \right)= 1$.
En cuyo caso,
$$ \sum_{s=2}^{\infty} \left( \zeta(s)-1 \right)= \sum_{s=2}^{\infty}\sum_{n=2}^{\infty} \frac{1}{n^{s}} = \sum_{n=2}^{\infty} \sum_{s=2}^{\infty} \frac{1}{n^{s}}$$
$$ = \sum_{n=2}^{\infty} \frac{\frac{1}{n^{2}}}{1-\frac{1}{n}} = \sum_{n=2}^{\infty} \frac{1}{n(n-1)}$$
$$=\sum_{n=2}^{\infty} \left( \frac{1}{n-1} - \frac{1}{n}\right) $$
$$ = \lim_{N \to \infty} \left(1- \frac{1}{2} + \frac{1}{2} -\frac{1}{3} + \ldots + \frac{1}{N-1} - \frac{1}{N} \right)$$
$$ = 1- \lim_{N \to \infty}\frac{1}{N} = 1$$