Is the given expression, monotonically increasing or decreasing with increasing x?


$\frac{1}{x \log(x)} \sum\_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$


EDIT: This is the derivative of the prime counting function $\pi(x)$ w.r.t. x, ie.,
$\frac{d\pi(x)}{dx} \sim \frac{1}{x \log(x)} \sum\_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$
Also, you might check Bruce C. Berndt's "Ramanujan's NoteBooks Part IV" page 123, equation 10.16, which gives = sign instead of $\sim$



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. 




---


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](http://mathworld.wolfram.com/GramSeries.html) 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.


