Is it a telescopic series?

From expansion series of $tan^{-1} x$ : $\tan^{-1} x=1-\dfrac{x^3}{3} +\dfrac{x^5}{5}- \dfrac{x^7}{7}....$ Put x=1 to get the required summation... $\Large\therefore~\dfrac{\pi}{4}=\sum _{ n=1 }^{ \infty }\frac { 1 }{ 4n-3 } -\frac { 1 }{ 4n-1 }$ $\Large A=4$ $\huge\therefore \dfrac{2016}{A}=\boxed{\color{#007fff}{504}}$ See this for further analysis of this series.

Expanding the summation (say S) and comparing it to the familiar Gregory series for $\arctan(x)$ , we get x=1

OR S = $\arctan(1)$ = $\frac{\pi}{4}$ .

Firstly We can write 1/4n-3 and 1/4n-1 as integral from 0 to 1 of x^(4n-4) and x^(4n-2).

Then interchanging summation and integration we will get two geometric progressions which on simplification will yield

Integration from 0 to 1 of 1/(1+x^2) which gives required result