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My question is if it's possible justify the following calculations, that are an application of Frullani's theorem in complex context on assumption of the Riemann Hypothesis to prove that $$\sum_{n=1}^\infty\sum_{k=0}^\infty(-1)^k\frac{2^{2k+1}((\gamma_n)^{2k+1}-(\gamma_{n+1})^{2k+1})}{2k+1}=\arctan(2\Im \rho_1),$$ where $\rho_n$ is the sequence of non-trivial zeros of the Riemann Zeta function with $\gamma_n:=\Im\rho_n>0$.

Question. Does makes sense $$\sum_{n=1}^\infty\sum_{k=0}^\infty(-1)^k\frac{2^{2k+1}((\gamma_n)^{2k+1}-(\gamma_{n+1})^{2k+1})}{2k+1}=\arctan(2\Im \rho_1)$$ on assumption of the Riemann Hypothesis? My purpose is understand those steps and where were my mistakes. Thanks in advance.

Notice that $$\sum_{k=0}^\infty\frac{(-1)^k(2a)^{2k+1}}{2k+1}=\arctan(2a)$$ converges only when $|a|<\frac{1}{2}$, but here $\gamma_1>14>\frac{1}{2}$, thus I don't know if previous series converges, this is, where was my mistake? I provide us a sketch of this exercise.

$\bullet$ Use Frullani Theorem in complex context to show that $$\sum_{n=1}^\infty\int_0^\infty\frac{e^{-x\rho_n}-e^{-x\rho_{n+1}}}{x}dx=-\mathcal{Log}(\rho_1),$$ notice that here could be a mistake because I've used the complex logarithm (main branch) property for quotients $\mathcal{Log}(\frac{b}{a})=\mathcal{Log}(b)-\mathcal{Log}(a)$ , to define a telescoping series in LHS. Was here my mistake?

$\bullet$ On assumption of the Riemann Hypothesis, take the imaginary part to get using the Taylor series for the sine function $$\sum_{n=1}^\infty\sum_{k=0}^\infty \left( \frac{(\gamma_{n})^{2k+1}-(\gamma_{n+1})^{2k+1}}{(2k+1)!}(-1)^k\int_0^\infty e^{-x/2}x^{2k}dx\right) =\arctan(2\Im \rho_1),$$ where we change the series with the integral sign using the Dominated Convergence Theorem.

Is right the swap the series and the integral sign? I believe that yes, since the function in the integral is bounded when one takes the complex modulus.

$\bullet$ Finally one uses the identity $$\int_0^\infty e^{-\frac{x}{2}}x^{2k}dx=2^{2k+1}\Gamma(2k+1),$$ for integers $k\geq 0$, which is right, you can use a CAS or integration by parts and mathematical induction.

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  • Sorry because were fixed some typos. –  Sep 09 '16 at 18:57
  • please, stop playing with formulas you don't understand (like completely divergent series). and $\int_0^\infty x^{2k} e^{-x/a} dx = a^{2k+1} \Gamma(2k+1)$ is a easy as $y = x/a$ (by the way, do you know how to prove the change of variable formula in general ? and the integration by parts ?) – reuns Sep 09 '16 at 22:11
  • Welcome @user1952009, thaks for your attention and mathematics. I know how prove $\int_0^\infty x^{2k} e^{-x/a} dx = a^{2k+1} \Gamma(2k+1)$ using integration by parts. If you can provide to me where was my mistake in this or previous question about RH I can understand. Then a comment should be very useful to me. –  Sep 09 '16 at 22:42
  • Which I am saying is that you are right (100%) about your method saying: one can do mathematics when one read and understand books, but it is hard. Also if you or other user tell me where is the mistake, well I try don't repeat it. Many thanks @user1952009 –  Sep 09 '16 at 22:49
  • you have the level for seing what is convergent and what is not (use $\rho_n \sim 1/2+i\frac{2\pi n}{\ln n}$). and if you can't prove the Frullani's theorem, don't use it – reuns Sep 09 '16 at 23:02
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    Let there be an analytic function given by the expansion $f(x)=\sum_{k=0}^\infty\frac{(-1)^k(2x)^{2k+1}}{2k+1}$ for $|a|<1/2$. The analytic continuation of $f(x)$ must then be given as $f(x)=\arctan(2x)$ for $x\in\mathbb C$ – Simply Beautiful Art Sep 10 '16 at 00:28
  • Yes @user1952009 I've read now the asymptotic for $|\gamma_n|$ as $\frac{2\pi n}{\log n}$, as $n$ tends to infinite. Thanks for your answer and reference. –  Sep 10 '16 at 07:37
  • and $2^{k} \frac{\gamma_n^{k}-\gamma_{n+1}^{k}}{k} \to \infty$ very fast (as $k \to \infty$) – reuns Sep 10 '16 at 08:02
  • Or just recall what the Riemann zeta hypothesis is to understand the nature of $\gamma$... – Simply Beautiful Art Sep 10 '16 at 13:06
  • Very thanks much for your useful comments @SimpleArt –  Sep 12 '16 at 16:41

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Say, $\sum_{n=0}^\infty(-1)^{n+1}-(-1)^n$ is telescoping, so surely it must equal $1$, the only term that does not cancel, which in other words says $$1=(-1+1)+(1-1)+(-1+1)+(1-1)+\dots$$How do you do telescoping unbounded partial sums? Obviously, if you assumed the Riemann Zeta hypothesis, then $\lim_{n\to\infty}\rho_n=\infty$, so you more or less just said

$$[\log(1)-\log(2)]+[\log(2)-\log(3)]+[\log(3)-\log(4)]\dots=\log(1)$$

just using different numbers is all you did.

So the real question is if you know what kind of math you are dealing with, because if you don't, you'll most certainly make mistakes, especially when it comes down to the different ways you can deal with certain things.

Note that

$$\sum_{n=1}^\infty a_n:=\lim_{N\to\infty}\sum_{n=1}^Na_n$$

which only exists if the limit exists.

if you did, however, have the background to understand the concept of $\sum_{n=1}^\infty a_n$ when the limit definition makes no sense, then you wouldn't ask if it made any sense because almost anyone who deals with such math does not attach "sense" to it. (yes, we are a little crazy, but I'm sure that's ok)

Simply Beautiful Art
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