# On computations related with $\lim_{x\to\infty} e^{-x}\sum_{\rho}\frac{(e^x)^\rho}{\rho}=0$

When I've reproduced the shape of the function $\sigma(x)$ of Apostol's section 4.10, a view of the page 98 is avaible as a Google Book (Apostol, Introduction to Analytic Number Theory, Springer 1976), in the LHS of the explicit formula (I say the formula in first paragraph of page 343, Conrey, The Riemann Hypothesis from Notices of the AMS) then the Prime Number Theorem implies $$\lim_{x\to\infty} e^{-x}\sum_{\rho}\frac{(e^x)^\rho}{\rho}=0,$$ where the series is over the non-trivial zeros of the Riemann Zeta function, and the nature/problems of such sum and series is explained in the subsequent paragraph of previous article. I believe that previous limit makes sense by my computations, but I don't know what is its full meaning, by the nature of such series.

Question 1. Is right previous limit? Is right the substitution that I've computed $x\leftarrow e^x$? Can you explain what is the meaning of such series and limit? It is, how we can understand previous limit? Thanks in advance.

Also I am interesting in the following, but if it is lot of work (you are welcome to answer it, if you want or well) provide us hints about the next

Question 2. Without the Prime Number Theorem what computations and statements can you prove about $$\lim_{x\to\infty} e^{-x}\sum_{\rho}\frac{(e^x)^\rho}{\rho}$$ when the series is interpreted as same as the cited paragraph? I say, that you can take the complex modulus, real and imaginary parts, get bounds, use the definition of limit...to obtain easy claims that I can learn without a heavy tool likes the Prime Number Theorem. Thanks in advance.

## 1 Answer

Let $$\psi\left(x\right)=\sum_{p^{m}\leq x}\log\left(p\right)$$ where $x$ is not a prime power. From the Perron's formula: we have $$\psi\left(x\right)=\int_{c-i\infty}^{c+i\infty}\sum_{p^{m}\leq x}\frac{\log\left(p\right)}{p^{ms}}\frac{x^{s}}{s}ds$$ $$=-\int_{c-i\infty}^{c+i\infty}\frac{\zeta'}{\zeta}\left(s\right)\frac{x^{s}}{s}ds$$ with $\textrm{Re}\left(s\right)>1$. Now if we move the contour to the left we have, from the residue theorem, that $$\psi\left(x\right)=x-\sum_{\rho}\frac{x^{\rho}}{\rho}-\frac{\zeta'}{\zeta}\left(0\right)-\frac{1}{2}\log\left(1-\frac{1}{x^{2}}\right)$$ where the first term follows from the pole of Zeta at $s=1$, the series from the non trivial zeros, $\frac{\zeta'}{\zeta}\left(0\right)$ from the pole at $s=0$ and $\frac{1}{2}\log\left(1-\frac{1}{x^{2}}\right)$ from the trivial zeros. This is essentially how we get the explicit formula. Note that the PNT is equivalent to $$\psi\left(x\right)\sim x$$ then $$x^{-1}\left|\sum_{\rho}\frac{x^{\rho}}{\rho}\right|\rightarrow0.$$ Note that the series converges conditionally but not absolutely, so there is no “easy claims”.

• Very thanks much, tomorrow at the morning I will and study your answer. I accept your words that there is no easy claims for the limit, I understand it. – user243301 Jun 8 '16 at 17:59
• @user243301 Actually I don't know if there are "easy claims" (I interpret it as easy manipulations of the series for proving convergence) but since there is no absolutely convergence I bet on the no. Note that if you take the sum considering each zeros with is conjugate then you have the absolutely convergence. – Marco Cantarini Jun 9 '16 at 10:27
• My main problem, thus I understand that your answer perfectly satisfactory, was is the substitution $e^x$ provide us some advantage in the knowledge, but now I understand that the limit is only the Prime Number THeorem. I thought in improve, in meaning, my question(s) (deleting some of my questions), but still I don't decide it . Thanks for your patience and attention. – user243301 Jun 9 '16 at 10:33
• @user243301 What I wrote holds for every positive (not prime power) $x$. So if we take $x=e^{y}$ the claim is the same. – Marco Cantarini Jun 9 '16 at 10:35
• Ok your last comment, with respect your previous question, as I've said in previous comments, my problem is that I am asking in this site by statements that I don't know very well. In previous myself Question, with easy claims about the limit, I ask if you or other user can deduce something about the limit $$\lim_{x\to\infty} e^{-x}\sum_{\rho}\frac{(e^x)^\rho}{\rho}$$ without great theorems, this is only with tools from a $1^{st}$ course in analysis. Very thanks much. – user243301 Jun 9 '16 at 10:39