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.

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.

true, or even mathematically meaningful. If you read Berndt's commentary, he makes it clear that this is not a true mathematical statement but one which is interesting from the viewpoint of giving an insight into Ramanujan's thought processes: "Of course Entry 8 is only a formal statement, but let us speculate how Ramanujan might have argued." – Pete L. Clark Jan 21 '10 at 4:04