TITLE: An infinite series involving the Zeta Function
QUESTION [9 upvotes]: I am wondering if anyone knows how to evaluate either of the following sums in terms of known constants:
$$\sum_{k=2}^{\infty}-\frac{\zeta^{'}(k)}{\zeta(k)},$$ 
and
$$\sum_{k=2}^{\infty}-\frac{\mu(k)}{k}\frac{\zeta^{'}(k)}{\zeta(k)}.$$ 
It is not hard to check that both converge absolutely, but based on (122)-(131) of the Wolfram Zeta Page, I think both sums should take on nice values.
References to any material which deals with these types of sums in general is also appreciated.
Thanks,

REPLY [9 votes]: Not really an answer to the question posed, but too big to be a comment.
The value of the first sum is
$$
  \sum_{k>=2} - \frac{\zeta^\prime(k)}{\zeta(k)} = 0.850312379764164578438788712404715501868902645375196564818394
$$
The above value is not recognized by Plouffe's inverter, unfortunately.
Moreover, because $\log \zeta(s) = - \sum_{k\ge1} \log (1-p_k^{-s})$ for $s>1$, it follows that
$$
  - \sum_{k>=2} \frac{\zeta^\prime(k)}{\zeta(k)} = \sum_{i \ge 1, k\ge 2}  \frac{\log p_i}{p_i^k -1}
$$
and even $\sum_{k>=1} (x^k-1)^{-1}$ is not known in closed form, so the chances are slim, but one never knows.
In regard to the other some, it comes close to 
$$
  \zeta_P(s) = \sum_{k \ge 1} p_k^{-s} = \sum_{n\ge 1} \frac{\mu(n)}{n} \log \zeta( s n) \qquad \text{ for } s > 1
$$
Differentiating with respect to $s$ and subtracting the pole term we would get
$$
 \lim_{s \to 1+} \zeta_P^\prime(s) - \frac{1}{1-s} = C + \sum_{k \ge 2} \mu(k) \frac{\zeta^\prime(k)}{\zeta(k)}
$$
which, again, is not quite the same. Numerical value for the second sum also does not turn up any results in Plouffe's inverter:
$$
  \sum_{k \ge 2} \frac{\mu(k)}{k} \frac{\zeta^\prime(k)}{\zeta(k)} =0.344146097673912783894171441679617569043972324522437879896534
$$
Why do you expect these sums to have nice values ?