Sum of Zetas Inequality

Question

I am here because I am unsure about how to prove the following: $$s-1 \geqslant \sum_{n=2}^{s-1}\zeta(n)\, \left\lbrace s\mid s\in\mathbb{Z^{+}},\, s\geqslant 2 \right\rbrace$$ where $\zeta(n)$ denotes the Riemann Zeta function. I believe it to be true, given the conditions, but I have not come up with a method sophisticated enough to prove such an inequality. Thank you.

Answer 1

Notice that $$ \begin{align} \sum_{k=2}^\infty(\zeta(k)-1) &=\sum_{k=2}^\infty\sum_{n=2}^\infty\frac1{n^k}\\ &=\sum_{n=2}^\infty\sum_{k=2}^\infty\frac1{n^k}\\ &=\sum_{n=2}^\infty\frac1{n^2\left(1-\frac1n\right)}\\ &=\sum_{n=2}^\infty\left(\frac1{n-1}-\frac1n\right)\\[6pt] &=1 \end{align} $$ Thus, for finite $s$, $$ \begin{align} 1 &\gt\sum_{n=2}^{s-1}(\zeta(k)-1)\\ &=\sum_{n=2}^{s-1}\zeta(k)-(s-2) \end{align} $$ Therefore, $$ s-1\gt\sum_{n=2}^{s-1}\zeta(k) $$