An amazing summation! (part-2)

$\displaystyle \sum_{n=1}^{\infty}\dfrac8{16n^2-16n+3} = \displaystyle \sum_{n=1}^{\infty}\dfrac8{(4n-1)(4n-3)}$

By turning it into partial fractions, we obtain:

$= \displaystyle 4\sum_{n=1}^{\infty}(\dfrac1{4n-3} - \dfrac1{4n-1})$

$=\large{4(\dfrac1{1} - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + \dfrac{1}{9} - \ldots \text { ad inf})}$

We know by Gregory's Series that:

$\large{\tan^{-1} (x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \ldots \text { ad inf}}$

Therefore, putting $x=1$ in the above series, we obtain the next step of solution as

$= 4\tan^{-1} (1) = 4(\dfrac{\pi}{4}) = \pi = \boxed{3.14}$