**Hint**. One may use standard Taylor series expansions as $i \to \infty$, to obtain $$ \begin{align} &(2i+4)\tan\left(\frac{\pi }{2i+4}\right)+(2i+3)\tan\left(\frac{\pi }{2i+3}\right)=2 \pi +\frac{\pi ^3}{6 i^2}+O\left(\frac1{i^3}\right) \\\\\\\ &\frac{\pi }{1-\frac{(i+2)\left(1-\sin\left(\frac{\pi (i+1)}{2i+4}\right)\right)^2}{2}}=\pi +O\left(\frac1{i^3}\right) \\\\\\\ &\frac{\pi }{1-(2i+3)\left(\frac{1-\sin\left(\frac{\pi (2i+1)}{2(2i+3)}\right)}{1+\sin\left(\frac{\pi (2i+1)}{2(2i+3)}\right)}\right)^2}=\pi +O\left(\frac1{i^3}\right) \end{align} $$ giving, as $i \to \infty$, a general term such that $$ u_i=\frac{\pi ^3}{6 i^2}+O\left(\frac1{i^3}\right) $$ and the given series is _convergent_ by the comparison test.