Is the following series ‘summable’ in the sense that it may be divergent but we can attach a meaning to it?

$$ \sum_{n=1}^{\infty} \frac{\Lambda(n)}{\sqrt{n}} \cos(x \log(n) + a) = (\text{reg}) \quad \frac{\zeta'(1/2 + ia + ix)}{\zeta(1/2 + ia + ix)} + \frac{\zeta'(1/2 - ia - ix)}{\zeta(1/2 - ia - ix)}. $$

If we assume that this is correct, then what would be the limit as $ x \to \infty $?

This series was obtained simply by setting $ s = 1/2 + ix $ inside the power series $$ \frac{\zeta'(s)}{\zeta(s)} = \sum_{n=1}^{\infty} \Lambda(n) n^{-s}. $$ Also, for large $ x $, this can be applied to the series $$ \sum_{n=1}^{\infty} \frac{\Lambda(n)}{\sqrt{n}} {J_{0}}(x \log(n)). $$

  • $\begingroup$ What are you trying to do here? $\endgroup$ – Ethan Feb 8 '13 at 0:31
  • $\begingroup$ i know the series is divergent... ETHAN however my question is if i can truncate it or give a 'sum' to this series for every 'x' $\endgroup$ – Jose Garcia Feb 12 '13 at 22:22

Your Answer

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

Browse other questions tagged or ask your own question.