I need Help to evaluate infinite series : $$S=1+\frac{1}{3^3}-\frac{1}{5^3}-\frac{1}{7^3}+\cdot\cdot\cdot$$
My try:
Let $$c_n:= \left({\frac{1-i}{2}}\right)(i)^n+\left({\frac{1+i}{2}}\right)(-i)^n$$
therfore:
$$S=\sum_{n=0}^{\infty} \frac{c_n}{(2n+1)^3}=\left({\frac{1-i}{2}}\right)\sum_{n=0}^{\infty}\frac{i^n}{(2n+1)^3} +\left({\frac{1+i}{2}}\right)\sum_{n=0}^{\infty}\frac{(-i)^n}{(2n+1)^3} $$
Is there a way to evaluate this series?
We can interpret this as a Dirichlet Series of the form. $$ \sum^{\infty}_{n=1}\frac{a_{n}}{n^s} = F(s) $$ We can then use the following therom of Mellin transforms of dirichlet series. $$ F(s) = \frac{1}{\Gamma(s)}\int^{\infty}_{0}x^{s-1}f(e^{-x})dx \; , \; f(z)=\sum^{\infty}_{n=1}a_{n}z^{n} $$ We can then find that $f(z)$ to be... $$ f(z) = \sum^{\infty}_{n=0}(-1)^{\frac{n(n-1)}{2}}z^{2n+1}=\frac{z(z^2+1)}{z^4+1}=\frac{1}{2\sqrt 2}\left(\frac{a}{a^*+z}+\frac{a^*}{a+z}-\frac{a}{a^*-z}-\frac{a^*}{a-z} \right) \; , \; a = \frac{1+i}{\sqrt 2} = e^{i\frac{\pi}{4}} $$ We can then find the general form that... $$ \frac{1}{\Gamma(s)}\int^{\infty}_{0}\frac{x^{s-1}}{\eta \pm e^{-x}}dx = \frac{1}{\eta\Gamma(s)}\int^{\infty}_{0} x^{s-1}dx + \frac{1}{\eta}Li_{s}\left(\frac{\pm 1}{\eta}\right) $$ Where $Li_{s}(x)$ is the Poly Log function. We can then combine this result to find... $$ F(s)= \frac{a}{a^* 2\sqrt 2}\left[ Li_{s}\left(-\frac{1}{a^*}\right) - Li_{s}\left(\frac{1}{a^*}\right) \right] + \frac{a^*}{a 2\sqrt 2}\left[ Li_{s}\left(-\frac{1}{a}\right) - Li_{s}\left(\frac{1}{a}\right) \right] $$ Since the Poly Log function is holomorphic this is the sum of complex conjugates hence we can simplify as the following. $$ F(s) = \frac{1}{\sqrt 2}\Im\left[Li_{s}\left( e^{i\frac{3}{4}\pi} \right) - Li_{s}\left( e^{i\frac{7}{4}\pi} \right)\right] $$ We can then get then seperate the real and complex components using the following identity... $$ Li_{s}\left(e^{2\pi i\frac{m}{p}}\right) = p^{-s}\sum^{p}_{k=1}e^{2\pi i\frac{m}{p}k} \zeta \left( s,\frac{k}{p} \right) $$ ... where $\zeta (s,a)$ is the Hurowitz-Zeta function. Thus... $$ F(s) = \frac{8^{-s}}{\sqrt 2}\sum^{8}_{k=1}\left(sin\left(\pi\frac{3}{4} k\right) - sin\left(\pi\frac{7}{4} k\right) \right) \zeta \left( s,\frac{k}{8} \right) = 8^{-s}\left( \zeta \left( s,\frac{1}{8} \right) + \zeta \left( s,\frac{3}{8} \right) - \zeta \left( s,\frac{5}{8} \right) - \zeta \left( s,\frac{7}{8} \right)\right) $$ We can then use the following reflection formula for the Hurowitz Zeta function... $$ (-1)^s \zeta(s,1-a)+\zeta(s,a) = \frac{\pi}{(s-1)!}\frac{d^{s-1}}{da^{s-1}} \cot(\pi a) $$ Assuming that s is an odd number... $$ F(s) = \frac{\pi}{(s-1)!8^s}\left[ \frac{d^{s-1}}{da^{s-1}} \cot(\pi a) \bigg{|}_{a=1/8} + \frac{d^{s-1}}{da^{s-1}} \cot(\pi a) \bigg{|}_{a=3/8} \right] $$ Hence for s = 3... $$ F(3) = \frac{3\pi^{3}}{64\sqrt 2} = 1.027722... $$
Doing the same as @GEdgar in comments, consider the partial sums $$S_p(a)=\sum_{n=0}^p \frac 1{(8n+a)^3}=\frac{\psi ^{(2)}\left(p+1+\frac{a}{8}\right)-\psi ^{(2)}\left(\frac{a}{8}\right)}{1024}$$ Compute $$A_p=S_p(1)+S_p(3)-S_p(5)-S_p(7)$$ and use the expansion $$\psi^{(2)}(q)=-\frac 1 {q^2}-\frac 1 {q^3}-\frac{1}{2 q^4}+O\left(\frac{1}{q^6}\right)$$
This would give $$A_p=\frac{-\psi ^{(2)}\left(\frac{1}{8}\right)-\psi ^{(2)}\left(\frac{3}{8}\right)+\psi ^{(2)}\left(\frac{5}{8}\right)+\psi ^{(2)}\left(\frac{7}{8}\right)}{1024 }-\frac{1}{512 p^3}+O\left(\frac{1}{p^4}\right)$$ Now, use
$$\psi ^{(2)}\left(\frac{7}{8}\right)-\psi ^{(2)}\left(\frac{1}{8}\right)=4 \left(4+3 \sqrt{2}\right) \pi ^3$$ $$\psi ^{(2)}\left(\frac{5}{8}\right)-\psi ^{(2)}\left(\frac{3}{8}\right)=4 \left(3 \sqrt{2}-4\right) \pi ^3$$
So $$A_p=\frac{3 \pi ^3}{64 \sqrt{2}}-\frac{1}{512 p^3}+O\left(\frac{1}{p^4}\right)$$
Edit for your curiosity
Using $$c_n=\sin \left(\frac{\pi n}{2}\right)+\cos \left(\frac{\pi n}{2}\right)$$ ans special function the result write also $$\frac {(1+i)}{16} \left(\Phi \left(-i,3,\frac{1}{2}\right)-i\, \Phi \left(i,3,\frac{1}{2}\right)\right)$$ where $\Phi(.)$ is the Lerch transcendent function.
This dessert was provided by my Facebook friend Noureddine Sma
we have : $$S=\sum_{n=0}^{\infty}\frac{(-1)^n}{(4n+1)^3}+\frac{(-1)^n}{(4n+3)^3}=\sum_{n=0}^{\infty}\frac{1}{(8n+1)^3}+\frac{-1}{(8n+5)^3}+\frac{1}{(8n+3)^3}+\frac{-1}{(8n+7)^3}$$
and we have $$\psi^{(2)}(z)=\sum_{n=0}^{\infty}\frac{-2}{(n+z)^3}$$
therfore: $$S=\frac{1}{2^{10}}(\psi^{(2)}(7/8)-\psi^{(2)}(1/8)+\psi^{(2)}(5/8)-\psi^{(2)}(3/8))$$
and using Identity: $$\psi^{(2)}(1-z)-\psi^{(2)}(z)=2\pi^3\csc^2(\pi z)\cot(\pi z)$$
therfore: $$S=\frac{2\pi^3}{2^{10}}(\csc^2(\pi/8)\cot(\pi /8)+\csc^2(3\pi/8)\cot(3\pi /8)=\frac{3\pi^3\sqrt2}{128}$$
