1729 followers problem (a bit late may be)

First note that $1-\dfrac{1+2x}{2} = \dfrac{1-2x}{2}$ .

Now we know the Euler's Reflection Formula : $\Gamma(n)\Gamma(1-n) = \dfrac{\pi}{\sin(n\pi)}$ . So in this problem we have $n=\dfrac{1+2x}{2}$ . So our summation changes to:

$\sum_{x=1}^{1729} \dfrac{\pi}{\sin\left(\frac{(2x+1)\pi}{2}\right)} = \pi\sum_{x=1}^{1728} \dfrac{1}{\sin\left(\frac{(2x+1)\pi}{2}\right)}+\dfrac{\pi}{\sin\left(\frac{3459\pi}{2}\right)}$

$=\pi\left[\dfrac{1}{\sin\dfrac{\pi}{2}}+\dfrac{1}{\sin\dfrac{3\pi}{2}}+\dfrac{1}{\sin\dfrac{5\pi}{2}}+\dfrac{1}{\sin\dfrac{7\pi}{2}}+\dots + \dfrac{1}{\sin\dfrac{3457\pi}{2}}\right] + \dfrac{\pi}{\sin\left(\frac{3\pi}{2}\right)}$

Now since $\sin(\pi+x)=-\sin(x)$ , we have $\sin\left(n\dfrac{\pi}{2}\right)=-\sin\left((n+2)\dfrac{\pi}{2}\right)$ and hence , $\left(\dfrac{1}{\sin\dfrac{\pi}{2}}+\dfrac{1}{\sin\dfrac{3\pi}{2}}\right)$ , $\left(\dfrac{1}{\sin\dfrac{5\pi}{2}}+\dfrac{1}{\sin\dfrac{7\pi}{2}}\right)$ , and so on till $\left(\dfrac{1}{\sin\dfrac{3455\pi}{2}}+\dfrac{1}{\sin\dfrac{3457\pi}{2}}\right)$ ALL EQUAL ZERO!. Hence we have $\sum_{x=1}^{1728} \dfrac{1}{\sin\left(\frac{(2x+1)\pi}{2}\right)} = 0$ .Also $\dfrac{\pi}{\sin\left(\frac{3\pi}{2}\right)} = -\pi$ , hence our summation becomes:

$=\pi \times 0 -\pi = -\pi\approx\boxed{-3.1415}$

Hint to Bonus Question: $1730$ is an even number.