I was looking at the following series $$S = \sum_{n=2}^\infty\frac{\sqrt{n}}{n(n-1)}$$ which WA says that it converges to some number (WA1: 2.18401, WA2: 2.0839...). I noticed that it is (almost) equal to $$\sum_{p=3,5,...}^\infty \zeta\left(\frac{p}{2}\right) = \zeta\left(\frac{3}{2}\right) + \zeta\left(\frac{5}{2}\right) + \cdots,$$ i.e., sum of "all halves" (from 3/2) of Rieman zeta functions.
Here is my thinking: $$\frac{\sqrt{n}}{n(n-1)} = \left(\frac{1}{n}\right)^\frac{3}{2}+\left(\frac{1}{n}\right)^\frac{5}{2}+\left(\frac{1}{n}\right)^\frac{7}{2}+\dots$$ Sum it over $n$ from 2 to $\infty$, you get e.g. for the first part $\sum_{n=2}^{\infty}\frac{1}{n^{3/2}},$ which is almost $\zeta\left(\frac{3}{2}\right)$, except that the summation does not start at $n=1$. To fix this, I do $\sum_{n=2}^\infty\frac{1}{n^{3/2}}=\zeta\left(\frac{3}{2}\right) - 1.$
Thus, I conclude that $$\sum_{p=3,5,...}^\infty \left(\zeta\left(\frac{p}{2}\right) - 1\right) = \sum_{n=1}^\infty\left(\zeta\left(n+\frac{1}{2}\right) - 1\right)= \sum_{n=2}^\infty\frac{\sqrt{n}}{n(n - 1)}.$$ It is also known (see https://en.wikipedia.org/wiki/Riemann_zeta_function) that $\sum_{n=2}^\infty\left(\zeta(n) - 1\right) = 1$. We can thus combine those sums, leading to $$\sum_{p=3}^\infty\left(\zeta\left(\frac{p}{2}\right) - 1\right)=\sum_{n=2}^\infty\frac{\sqrt{n}}{n(n-1)} + 1$$
My question is. If it is correct, is this identity known? If so, can we conclude that $\sum_{n=2}^\infty\frac{\sqrt{n}}{n(n-1)}$ is an irrational number (sum of square roots "feels" like an irrational number) and thus also the sum of "Riemann halves" is irrational number?
Also, is there something known about the $\sum_{p=1,3,5,...}^\infty \zeta\left(\frac{p}{2}\right)$? Its convergence, etc.
Thanks.
Answer to the question if it is known: It seems that the same equality has been mentioned lately in The Spiral of Theodorus and Sums ofZeta-values at the Half-integers