TITLE: inequality $ \prod_{n=2}^{\infty}n^{\zeta(n)-1} <\frac{\pi^2+6}{6}$
QUESTION [7 upvotes]: Let $ \zeta(s) $be the riemann zeta function, then

$$ \prod_{n=2}^{\infty}n^{\zeta(n)-1} <1+\frac{\pi^2}{6}$$


The problem is difficult, I don't know how to go started
Thank you very much!

REPLY [2 votes]: This product is easy as $\pi$. Take its logarithm, use the fact that the logarithm of a product is the sum of logarithms, then the fact that $\ln n^a=a\ln n$, with $a=\zeta(n)-1$. Now express a as an infinite series, and switch the order of the two sums, using various identities related to polylogarithms, the Riemann $\zeta$ and Dirichlet $\eta$ functions. Then, exponentiating, we get the inequality that you wanted to write all along.
$$\ln\prod_{n=2}^\infty n^{\zeta(n)-1}=\sum_{n=2}^\infty\ln n^{\zeta(n)-1}=\sum_{n=2}^\infty(\ln n)\cdot\Big(\zeta(n)-1\Big)=\sum_{n=2}^\infty\ln n\cdot\sum_{k=2}^\infty\frac1{k^n}=$$ $$=\sum_{k=2}^\infty\sum_{n=2}^\infty\frac{\ln n}{k^n}<\sum_{k=2}^\infty\sum_{n=2}^\infty\frac n{k^n}=\sum_{k=2}^\infty\frac{2-\tfrac1k}{(k-1)^2}=1+\zeta(2).$$