Analysis of series containing Zeta function:

Question

Consider the given alternating series:

$$f(x) =2\sum_{n=0}^\infty \frac{a_n(x-1)^{2n+1}}{\zeta(-2n-1)}$$

Here, $$a_n= \left(1+\frac{1}{4n+1}\right)^6\left(1-\frac{1}{4n+3}\right)^6\left(\frac{4n^2}{4n^2-1}\right)^2.$$

How can we calculate the asymptotic of this series as $x\rightarrow\infty$?

Atleast how to compute values of this series for atleast upto $x=100$ accurately?

Answer 1

To begin, using the formula for the zeta function for even integers, we have the following:$$\frac{a_n(2\pi)^{2n+2}(-1)^{n+1}}{\zeta(2n+2)\Gamma(2n+2)}=-a_n(2\pi)^{2n+2}\cdot\frac{2(2n+2)!}{(2\pi)^{2n+2}B_{2n+2}}\frac{1}{(2n+1)!}=-\frac{4n+4}{B_{2n+2}}a_n,$$

so that

$$f(x)=-4\sum_{n\ge0}\frac{n+1}{B_{2n+2}}a_n(x-1)^{2n+1}.$$

Answer 2

Not a full answer. Define $$ F(x): = - \frac{{f(\sqrt x + 1)}}{{2\sqrt {\pi x} }} = \sum\limits_{n = 0}^\infty {\frac{{a_n \pi ^{2n + 2} }}{{\zeta (2n + 2)\Gamma\! \left( {n + \frac{3}{2}} \right)}}\frac{{( - x)^n }}{{n!}}} . $$ Then, by Ramanujan's master theorem, $$ \int_0^{ + \infty } {x^{s - 1} F(x)\,\mathrm{d}x} = \frac{{a_{ - s} \pi ^{2 - 2s} \Gamma (s)}}{{\zeta (2 - 2s)\Gamma \!\left( {\frac{3}{2} - s} \right)}}. $$ Using the Mellin inversion formula, we then obtain $$ F(x) = \frac{1}{{2\pi \mathrm{i}}}\int_{c - \mathrm{i}\infty }^{c + \mathrm{i}\infty } {x^{ - s} \frac{{a_{ - s} \pi ^{2 - 2s} \Gamma (s)}}{{\zeta (2 - 2s)\Gamma\! \left( {\frac{3}{2} - s} \right)}}\mathrm{d}s} $$ with any $0 < c < \frac{1}{2}$. Thus, $$ f(x) = \frac{\mathrm{i}}{{\sqrt \pi }}\int_{c - \mathrm{i}\infty }^{c + \mathrm{i}\infty } {(x - 1)^{1 - 2s} \frac{{a_{ - s} \pi ^{2 - 2s} \Gamma (s)}}{{\zeta (2 - 2s)\Gamma\! \left( {\frac{3}{2} - s} \right)}}\mathrm{d}s} . $$ You may then deform the contour and, depending on the location of the non-trivial zeros of the Riemann zeta function, obtain some estimates for $f(x)$. A crude esimate under the Riemann hypothesis would be $f(x) = \mathcal{O}(x^{ - 1/2 + \varepsilon } )$.