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Let $\Lambda(n)$ be the Von-Mangoldt function which appears in the theory of zeta-function.

We know that \[ \sum_{k \geq 1} \frac{\Lambda(k)}{k^{\sigma + it}} = - \frac{\zeta'}{\zeta}(\sigma + it), \] and for any fixed $\sigma < 0$, $\frac{\zeta'}{\zeta}(\sigma + it)$ grows at most logarithmically.

Do we know anything maybe similar to this about the order of \[ F(\sigma + it) = \sum_{k \geq 2} \frac{\Lambda(k - 1)}{k^{\sigma + it}} \] as $t \to \infty$?

At first sight, I thought this is an easy problem.

Here is what I did.

I wrote with Abel summation formula, \[ F(s) = \sum_{k \geq 2} \frac{\Lambda(k - 1)}{k^{\sigma + it}} = s \int_{1}^{\infty} \frac{\sum_{i \leq t} \Lambda(i - 1) }{v^{s + 1}} dv \] But \[ \sum_{i \leq t} \Lambda(i - 1) - \sum_{i \leq t} \Lambda(i) = -\Lambda( \lfloor t \rfloor ) \] if $\Lambda(\lfloor t \rfloor ) \not= 0$ and otherwise it is zero.

Thus, \[ F(s) = - \frac{\zeta'}{\zeta}(s) - s \sum_{k \geq 1} \Lambda(k)\int_{k}^{k + 1}v^{-s - 1} dv. \]

The sum is convergent for $\sigma > 0$, but the factor $s$ is too strong, just as we suffer in any other situation.

It doesn't help with the negative $\sigma$.

Is there any method suitable for dealing with this?

Grown pains
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  • $(k+1)^{-s} =k^{-s}(1+\frac1{k})^{-s}= \sum_{m \ge 0} k^{-s-m} {-s \choose m}=\sum_{m =0}^{M-1} k^{-s-m} {-s \choose m}+ \mathcal{O}(k^{-s-M})$ thus $\sum_{k=1}^\infty \Lambda(k) (k+1)^{-s} = \sum_{m =0}^{1-\Re(s)} \frac{\zeta'(s+m)}{\zeta(s+m)}{-s \choose m} + \mathcal{O}(1)$ – reuns Jun 11 '17 at 11:25
  • And you meant $\sigma > 1$ or the analytic continuation of $\sum_{k=1}^\infty \Lambda(k) k^{-s},\sum_{k=1}^\infty \Lambda(k) (k+1)^{-s}$ – reuns Jun 11 '17 at 11:30
  • I meant the analytic continuation beyond $\sigma = 0$. Shouldn't there be a factor $\Im(s)$ inside the big-oh? – Grown pains Jun 11 '17 at 21:36
  • $\binom{s}{m} = \frac{s(s - 1) \cdots (s - m + 1)}{m!} \asymp \Im(s)^{m}$. Thus your expansion doesn't help at all. – Grown pains Jun 12 '17 at 21:33
  • Of course it helps It is not because $\sum_{k=1}^\infty \Lambda(k) k^{-s}$ and $\sum_{k=1}^\infty \Lambda(k) (k+1)^{-s}$ are close on $\Re(s) > 1$ that they should stay close on $\Re(s) \le 1$ – reuns Jun 12 '17 at 21:48
  • $(k + 1)^{-s} = k^{-s} (1 + 1/k)^{-s} = \sum_{m \geq 0}\binom{-s}{m} k^{-s - m}$, and summing over all positive integers $k \geq 1$, we get $\zeta(s) - 1 = \sum_{m \geq 0} \binom{-s}{m} \zeta(s + m) = \zeta(s) + \sum_{m \geq 1} \binom{-s}{m} \zeta(s + m) $. Thus, it is correct that the error term $\sum_{m \geq 1} \binom{-s}{m} \zeta(s + m)$ is very small in any vertical strip for this case. Is there any simple cancellation or rewriting in the error term that I don't recognize? – Grown pains Jun 21 '17 at 21:59

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