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

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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.

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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”.