What will be the sum?

$\textrm{I found this interesting}$

$\mathrm{tanx = \frac{1}{\frac{π}{2}-x} -\frac{1}{\frac{π}{2}+x} +\frac{1}{\frac{3π}{2}-x}-\frac{1}{\frac{3π}{2}+x} + \frac{1}{\frac{5π}{2}-x } -\cdots}$

$\textrm{ This lemma is a consequence of "Euler's product" formula}$

$\textrm{taking}$ $\mathrm{x= \frac{π}{6}}$

$\textrm{We will end up with}$

$\mathrm{ tan(\frac{π}{6}) = \frac{6}{2π} -\frac{6}{4π} + \frac{6}{8π} - \frac{6}{10π} + \frac{6}{14π} -\cdots}$

$\textrm{So}$ $\mathrm{\frac{π}{6√3} = \frac{1}{2} -\frac{1}{4} +\frac{1}{8} -\frac{1}{10} + \frac{1}{14} -\frac{1}{16} +\cdots }$

$\textrm{By Adding}$ $\frac{1}{2}$ $\textrm{we can get our answer}$

$\textrm{Answer will be \boxed { \frac{π}{6√3} +\frac{1}{2} } }$

The above series can be written as:

$1 + \Sigma_{n=0}^{\infty} \frac{1}{6n+8} - \frac{1}{6n+4};$

or $1 + \Sigma_{n=0}^{\infty} \int_{0}^{1} x^{6n+7} - x^{6n+3} dx$ ;

or $1 + \int_{0}^{1} (x^7-x^3) \cdot \Sigma_{n=0}^{\infty} x^{6n} dx;$

or $1 + \int_{0}^{1} \frac{x^7-x^3}{1-x^6} dx$ ;

or $1 - \int_{0}^{1} \frac{x^3(x^4-1)}{(x^3+1)(x^3-1)} dx;$

or $1 - \int_{0}^{1} \frac{x^3(x^2+1)(x+1)(x-1)}{(x+1)(x^2-x+1)(x-1)(x^2+x+1)} dx$ ;

or $1 - \int_{0}^{1} \frac{x^5+x^3}{x^4+x^2+1} dx$ ;

or $1 - \int_{0}^{1} x - \frac{x}{x^4+x^2+1} dx$ ;

or $1 - \int_{0}^{1} x - \frac{x}{(x^2+1/2)^2 + 3/4} dx$ ;

or $1 - [\frac{x^2}{2} - \frac{1}{\sqrt{3}} \arctan (\frac{2x^2+1}{\sqrt{3}})|_{0}^{1}];$

or $\boxed{\frac{1}{2} + \frac{\pi}{6\sqrt{3}}}.$