# On the series $\displaystyle \sum_{n=1}^{\infty} \frac{\Lambda(n)}{\sqrt{n}} \cos(x \log(n) + a)$.

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 $$\sum_{n=1}^{\infty} \frac{\Lambda(n)}{\sqrt{n}} {J_{0}}(x \log(n)).$$

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What are you trying to do here? – Ethan Feb 8 '13 at 0:31
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' – Jose Garcia Feb 12 '13 at 22:22