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?