I've deduced this identity involving the Riemann Zeta function $\zeta(s)$ and the Möbius function $\mu(n)$, the zeros and the meaning of how is understood the summation $\sum_{\rho}$ on assumption of the Riemann Hypothesis is explained in [2]. See my details below of this question.
Question 1. Is right that this identity $$\sum_{k=1}^\infty\frac{1}{k(k+1)}\sum_{\rho}\frac{k^\rho}{\rho\zeta'(\rho)}+\sum_{l=1}^\infty\frac{(-1)^{l-1}(2\pi)^{2l}}{(2l)!l\zeta(2l+1)}\sum_{k=1}^\infty \frac{1}{(k+1)k^{2l+1}}=2+\frac{1}{2}\sum_{k=1}^\infty\frac{\mu(k)}{k+1}$$ holds? Thanks in advance.
My deduction was to combine the identity that satisfy the Mertens and Möbius functions, that is (1.8) of page 2 from [1] (currently available in arXiv with article ID arXiv:math/0011254) with the exact formula of the Mertens function, see [2], I am saying the identity (2.3) in page 141, thus we are on assumption of the Riemann Hypothesis and the summation $\sum_{\rho}$ has the meaning (but I employ previous notation) explained by the authors. The simplifications were using the Prime Number Theorem (I know also that RH implies PNT), then from $$-\frac{1}{2}\sum_{k=1}^{n-1}\frac{\mu(k)}{k(k+1)}+\sum_{k=1}^{n-1}\frac{M(k)}{k(k+1)}=\sum_{k=1}^{n-1}\frac{1}{k(k+1)}\left(\sum_{\rho}\frac{k^\rho}{\rho\zeta'(\rho)}\right)-2\sum_{k=1}^{n-1}\frac{1}{k(k+1)}$$ $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\sum_{k=1}^{n-1}\frac{1}{k(k+1)}\sum_{l=1}^\infty \frac{(-1)^{l-1}(2\pi)^{2l}}{(2l)!l\zeta(2l+1)k^{2l}} ,$$
I've calculated using the identity that satisfy the Mertens and Möbius functions with the Prime Number Theorem that $$-\frac{1}{2}\left(0-\sum_{k=1}^\infty\frac{\mu(k)}{k+1}\right)+0-0=\sum_{k=1}^\infty\frac{1}{k(k+1)}\left(\sum_{\rho}\frac{k^\rho}{\rho\zeta'(\rho)}\right)-2$$ $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\sum_{l=1}^\infty\frac{(-1)^{l-1}(2\pi)^{2l}}{(2l)!l\zeta(2l+1)}\sum_{k=1}^\infty \frac{1}{(k+1)k^{2l+1}}.$$
Question 2. A) Is it possible identify the series $$\sum_{k=1}^\infty \frac{1}{(k+1)k^{2l+1}}?$$ B) (Optional, but I would like to see your answer.) Does make sense (does converge) this expression $$\sum_{\rho}\frac{1}{\rho\zeta'(\rho)}\sum_{k=1}^\infty\frac{k^\rho}{k(k+1)},$$ where $\sum_{\rho}$ means the cited limit? Many thanks.
With Wolfram Alpha I can get some examples of A) for simple values of $l$. If you know how derive a closed form or do you know from the literature please tell/refers me. For B) I know the calculation $$ \left| \sum_{k=1}^\infty\frac{k^\rho}{k(k+1)}\right| \leq \sum_{k=1}^\infty\frac{k^{\Re s}}{k(k+1)}=\sum_{k=1}^\infty\frac{1}{k^{1-\Re s}(k+1)}$$ that is convergent if $\Re s<1$, but I don't know if it is possible/known what about the series in Question 2.B).
References:
[1] Báez-Duarte, Arithmetical Aspects of Beurling's Real Variable Reformulation of the Riemann Hypothesis, (2000).
[2] Odlyzko and te Riele, Disproof of the Mertens conjecture, Journal für die reine und angewandte Mathematik (1985)
For understanding $\sum_\rho$ (sum over the non-trivial zeros) you need to study the analytic properties of $\zeta(s)$ (analytic continuation, functional equation, growth rate) the residue theorem and possibly some complicated estimations for the growth rate of $1/\zeta(s)$ and the density of zeros on the critical strip (on the critical line if assuming RH).
Everything is explained in the books on the Riemann zeta function : Titchmarsh, Apostol, Edwards, Montgomery.. There are also some teachers notes like thisone proving in details the prime number theorem for $\zeta(s)$ and $L(s,\chi)$ (in arithmetic progressions).