Summation of square root sine Is it possible to solve this?
$$y = \sum_{N=1}^{x}\sqrt{\sin N}$$
I know it is possible to solve $$y = \sum_{N=1}^{x}{\sin N}$$ by expressing sinN as its exponential before doing geometric progression to solve it. 
But unlike the above,$$\sqrt{\sin N}$$ cannot be converted to its exponential as $$Im[e^{i\frac{N}{2}}] 
= i\sin\frac{N}{2}.$$
Edited:
sinN is always positive as 0
 A: Rewrite sine using the complex definition:
$$\sum_{n=0}^{x}\sqrt{\sin(n)}=\sum_{n=0}^{x}\sqrt{\frac{e^{ni}-e^{-ni}}{2i}}=\frac1{\sqrt{2i}}\sum_{n=0}^{x}\sqrt{e^{ni}-e^{-ni}}$$
Perform binomial expansion.
$$\frac1{\sqrt{2i}}\sum_{n=0}^{x}\sqrt{e^{ni}-e^{-ni}}=\frac1{\sqrt{2i}}\sum_{n=0}^{x}\sum_{k=0}^{\infty}\frac{\Gamma(3/2)}{k!\Gamma(3/2-k)}(-1)^ke^{(1/2-k)ni}e^{-kni}$$
$$=\frac1{\sqrt{2i}}\sum_{n=0}^{x}\sum_{k=0}^{\infty}\frac{\Gamma(3/2)}{k!\Gamma(3/2-k)}(-1)^ke^{(1/2-2k)ni}$$
$$=\frac1{\sqrt{2i}}\sum_{k=0}^{\infty}\sum_{n=0}^{x}\frac{\Gamma(3/2)}{k!\Gamma(3/2-k)}(-1)^ke^{(1/2-2k)ni}$$
$$=\frac1{\sqrt{2i}}\sum_{k=0}^{\infty}\frac{\Gamma(3/2)}{k!\Gamma(3/2-k)}(-1)^k\sum_{n=0}^{x}e^{(1/2-2k)ni}$$
$$=\frac1{\sqrt{2i}}\sum_{k=0}^{\infty}\frac{\Gamma(3/2)}{k!\Gamma(3/2-k)}\frac{(-1)^k(1-e^{((1/2-2k)i)(x+1)})}{1-e^{(1/2-2k)i}}$$
If it comes to your attention that someone could get the last line into the following form:
$$=\frac1{\sqrt{2i}}\sum_{k=0}^{\infty}\frac{\Gamma(3/2)}{k!\Gamma(3/2-k)}a^{1/2-k}b^k$$
Then it would equal $(a+b)^{1/2}$ and that would solve everything.  My knowledge on summations does not let me directly solve the current form so I am somewhat stuck on helping you any further.
