The Appropriate Identity

Let us rewrite the summation in consideration as follows:

$\displaystyle\sum\limits_{k=0}^{\infty} \mathrm{arctan} \big( \frac{ 4}{4k^2 + 3} \big) = \mathrm{arctan}\big( \frac{4}{3} \big) + \displaystyle\sum\limits_{k=1}^{\infty} \mathrm{arctan} \big( \frac{ 4}{4k^2 + 3} \big)$

We could manipulate the RHS as follows:

$= \mathrm{arctan}\big( \frac{4}{3} \big) + \displaystyle\sum\limits_{k=1}^{\infty} \mathrm{arctan} \bigg( \frac{ 1}{k^2 + \frac{3}{4}} \bigg)$

$= \mathrm{arctan}\big( \frac{4}{3} \big) + \displaystyle\sum\limits_{k=1}^{\infty} \mathrm{arctan} \bigg( \frac{ 1}{k^2 - \frac{1}{4} + 1} \bigg)$

$= \mathrm{arctan}\big( \frac{4}{3} \big) + \displaystyle\sum\limits_{k=1}^{\infty} \mathrm{arctan} \bigg( \frac{ 1}{[k - \frac{1}{2}][k + \frac{1}{2}] + 1} \bigg)$

$= \mathrm{arctan}\big( \frac{4}{3} \big) + \displaystyle\sum\limits_{k=1}^{\infty} \mathrm{arctan} \bigg( \frac{ [k + \frac{1}{2}] - [k- \frac{1}{2}]}{[k - \frac{1}{2}][k + \frac{1}{2}] + 1} \bigg)$

We can now use the sum/difference identity for arctangents, which is as follows:

$\mathrm{arctan}(A) - \mathrm{arctan}(B) = \mathrm{arctan}\bigg( \frac{A-B}{1+ AB} \bigg)$

Since we have $A = k + \frac{1}{2}$ and $B = k - \frac{1}{2}$ , we further simplify the RHS as

$= \mathrm{arctan}\big( \frac{4}{3} \big) + \displaystyle\sum\limits_{k=1}^{\infty} \bigg[ \mathrm{arctan} \big( k + \frac{1}{2} \big) - \mathrm{arctan} \big( k - \frac{1}{2} \big) \bigg]$

which, when evaluated, turns out to be a telescoping series.

This will further lead us to

$= \mathrm{arctan}\big( \frac{4}{3} \big) + \frac{ \pi}{2} - \mathrm{arctan} \big( \frac{1}{2} \big)$

$\approx 2.03444$

Thus giving us the answer to our question, $\boxed {2034}$ .