Problem - H. Ohtsuka (Japan)
For $s>1$, prove the following inequalities:
\[\sum_p\frac{1}{p^s-\frac{1}{2}}<\log(\zeta(s)),\,\sum_p\frac{1}{p^s}<\log\left(\frac{\zeta(s)}{\sqrt{\zeta(2s)}}\right),\,\sum_p\frac{1}{p^s+\frac{1}{2}}<\log\left(\frac{\zeta(s)}{\zeta(2s)}\right)\]
where all sums are taken over prime numbers $p$, and $\zeta(s)=\sum_{k=1}^\infty1/k^s$.
Solution
We make use of the Euler product $\zeta(s)=\prod_p\frac{1}{1-1/p^s}$, $s>1$, and the Mercator series $\log(1+x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}x^n$, $-1<x<1$, for all three inequalities.
1)
\[\begin{align*} \log(\zeta(s)) &= -\sum_p\log\left(1-\frac{1}{p^s}\right)\\ &= \sum_p\sum_{n=1}^\infty\frac{1}{np^{ns}}\\ &>\sum_p\sum_{n=1}^\infty\frac{1}{2^{n-1}p^{ns}}\\ &=\sum_p\frac{2}{2p^s-1}\\ &=\sum_p\frac{1}{p^s-\frac{1}{2}}.\end{align*}\]
2)
\[\begin{align*} \log\left(\frac{\zeta(s)}{\sqrt{\zeta(2s)}}\right) &= \sum_p\left(-\log\left(1-\frac{1}{p^s}\right)+\frac{1}{2}\log\left(1-\frac{1}{p^{2s}}\right)\right)\\ &= \sum_p\left(-\log\left(1-\frac{1}{p^s}\right)+\frac{1}{2}\log\left(\left(1-\frac{1}{p^{s}}\right)\left(1+\frac{1}{p^{s}}\right)\right)\right)\\ &=\frac{1}{2}\sum_p\left(\log\left(1+\frac{1}{p^s}\right)-\log\left(1-\frac{1}{p^s}\right)\right)\\ &=\frac{1}{2}\sum_p\left(\sum_{n=1}^\infty\frac{(-1)^{n+1}}{np^{ns}}+\sum_{n=1}^\infty\frac{1}{np^{ns}}\right)\\ &= \frac{1}{2}\sum_p\sum_{n=0}^\infty\frac{2}{(2n+1)p^{(2n+1)s}}\\ &>\sum_p\frac{1}{p^s}.\end{align*}\]
3)
\[\begin{align*} \log\left(\frac{\zeta(s)}{\zeta(2s)}\right) &= \sum_p\left(-\log\left(1-\frac{1}{p^s}\right)+\log\left(1-\frac{1}{p^{2s}}\right)\right)\\ &= \sum_p\left(\log\left(\left(1-\frac{1}{p^{s}}\right)\left(1+\frac{1}{p^{s}}\right)\right)-\log\left(1-\frac{1}{p^s}\right)\right)\\ &= \sum_p\log\left(1+\frac{1}{p^{s}}\right)\\ &=\sum_p\sum_{n=1}^\infty\frac{(-1)^{n+1}}{np^{ns}}\\ &>\sum_p\sum_{n=1}^\infty\frac{(-1)^{n+1}}{2^{n-1}p^{ns}}\\ &=\sum_p\frac{2}{2p^s+1}\\ &=\sum_p\frac{1}{p^s+\frac{1}{2}}.\end{align*}\]
