About Riemann-Zeta function

$\zeta(1) = + \infty$ because $\zeta(1)$ is the harmonic series. Furthemore, $1 < \zeta(k + 1) < \zeta(k), \space \forall k \in \mathbb{N}( - \{0\})$ and $\displaystyle \lim_{k\to\infty} \zeta(k) = 1 \neq 0$ . Therefore, the above product diverges.

It is a standard result about the Riemann zeta function that $\sum_{k=2}^\infty \big[\zeta(k)-1\big] \; = \; 1$ and so we can write $\zeta(k) = 1 + a_k$ where $a_k > 0$ for all $k \ge 2$ and $\sum_{k=2}^\infty a_k = 1$ Thus we note that the sequence of finite products $\prod_{k=2}^K \zeta(k) \; =\; \prod_{k=2}^K\big(1 + a_k\big) \hspace{2cm} k \ge 2$ is increasing and positive, and $\prod_{k=2}^K \zeta(k) \; \le \; \prod_{k=2}^K e^{a_k} \; = \; \mathrm{exp}\left(\sum_{k=2}^K a_k\right) \; \le \; e^1 \; = \; e$ for all $K \ge 2$ . Thus the sequence of partial products is increasing and bounded above by $e$ , and hence we deduce that the infinite product converges and $\prod_{k=2}^\infty \zeta(k) \; \le \; e$