nt.number theory - Is the given expression, monotonically increasing or decreasing with increasing x?

I'm not sure I should bother answering this question, because it seems like the original poster may not have asked the right question. However, it is a nice exercise in basic asymptotics.

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For $x$ sufficiently large, the sum in question is decreasing.

First, note that this sum is equal to
$$sum_{k geq 1} frac{(log x)^{k-1}}{x k! zeta(k+1)}.$$
(See here for a very similar series; we are using the highly nontrivial identity $sum mu(i)/i=0$ to get rid of the "$k=0$" term.)

Substituting $x=e^u$, we want to know whether or not
$$e^{-u} sum_{k geq 1} frac{u^{k-1}}{k! zeta(k+1)}$$
is increasing or decreasing in $u$. One can justify taking term by term derivatives, so we want to know whether
$$e^{-u} left( sum_{k geq 2} frac{(k-1) u^{k-2}}{k! zeta(k+1)} - sum_{k geq 1} frac{u^{k-1}}{k! zeta(k+1)} right)$$
is positive or negative.

Rearranging terms, we are interested in the sign of
$$e^{-u} sum_{ell geq 0} frac{u^{ell}}{ell!} left( frac{ 1}{(ell+2) zeta(ell+3)} - frac{1}{(ell+1) zeta(ell+2)} right).$$

The quantity in parenthesis is $-1/(ell+1)(ell+2) + O(2^{- ell})$. So we are interested in the sign of
$$e^{-u} left( - sum_{ell geq 0} frac{u^{ell}}{(ell+2)!} + sum O(frac{ u^{ell} 2^{- ell}}{ell !}) right) =$$
$$e^{-u} left( - frac{e^u -1-u}{u^2} + O(e^{u/2}) right)=$$
$$-1/u^2 + O(e^{-u/2}).$$

This is negative for $u$ sufficiently large.