9
$\begingroup$

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,

$\endgroup$
2
  • $\begingroup$ If you can calculate a few decimals, there are places to look things up. $\endgroup$ Aug 18 '11 at 1:24
  • $\begingroup$ None of (122)-(131) involve $\zeta'$, so it's not clear to me why those formulas lead you to think that sums with $\zeta'$ will be "nice". $\endgroup$ Aug 18 '11 at 3:56
9
$\begingroup$

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 ?

$\endgroup$
1
  • $\begingroup$ +1, I like this answer. However the plouffe inverter is not very good at recognizing constants from number theory. For example, $B_3$ in equ (17) on this page:mathworld.wolfram.com/MertensConstant.html this constant does come up in a few places. $\endgroup$ Aug 18 '11 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.