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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 $\sim$

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If you asked someone this in the flesh, would you not normally explain what you'd tried already, or where the question came from? -1 for lack of context, or indication of your train of prior thought –  Yemon Choi Jan 21 '10 at 3:12
Um, unless I'm missing something here, isn't the derivative of $\pi(x)$ equal to 0 for everything except prime numbers, and undefined at primes themselves? –  Zev Chonoles Jan 21 '10 at 3:41
rpg16, only Ramanujan can get away with that. –  Jonas Meyer Jan 21 '10 at 3:59
@rpg16: Ramanujan's Notebooks are preserved for posterity because there are hundreds or thousands of intriguing formulas and conjectures in them. However, this is not to say that everything that appears in his notebooks is 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
@Roupam Ghosh: It is fine to be a learner. However, some of your behavior is alienating the research community: (i) not taking time to understand when people point out errors in your reasoning. [(i)' You have asked questions containing significantly flawed reasoning, which has been pointed out and corrected by several MOers, including me. You have never upvoted an answer, accepted an answer, or even clearly acknowledged your mistakes.] (ii) You have posted 7 versions of a paper to the arxiv, most recently tonight, on a subject you have not mastered. This is not what the arxiv is for. –  Pete L. Clark Jan 21 '10 at 4:18

3 Answers 3

up vote 8 down vote accepted

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.

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Thanks David... –  Roupam Ghosh Jan 21 '10 at 18:26

The expression $\pi(x)\approx\sum_{n>0}\frac{\mu(n)}{n}Li(x^{1/n})$ is certainly not an exact equality of any kind, and must be interpreted very carefully. This infinite series is sometimes called the Gram series; the real interest in this expression is that numerically it seems to have a much better error than the simple approximation $\pi(x)=Li(x)+error$. Ramanujan was misled by its remarkable accuracy (for smallish values of $x$) and made some very strong conjectures on the relation between $\pi(x)$ and the Gram series, conjectures which are demonstrably false. In fact Littlewood proved that the Gram series is a worse approximation to $\pi(x)$ than $Li(x)$ infinitely often.

Besides, $\pi(x)$ is a step function; it doesn't make any sense to differentiate a continuous approximation and expect to retain any meaning. Consider, for example, what would happen if you differentiated the expression $\left\lfloor x \right\rfloor\approx x+\sin(10^{10}x)$...

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In fact, if you plot the Ramanujan's series in Mathematica, up to i=1000, or so, what you get is a monotonically decreasing function. This function is indistinguishable of the derivative of the Riemann's prime counting function, R(x).

In other words, if there exists a function that match the prime counting function Pi(x), for all x (not for the steps, but for the mean behaviour), its derivative should be the Ramanujan's series.

The Ramanujan's series for the derivative of Pi(x) is a continuous function.

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