How to prove that this infinite series converge

Hello,

Can someone help me to prove that this infinite series converge / diverge ?
I've been trying for hours, and still no luck

http://pluspi.org/wiki/images/math/f/3/d/f3d91a7bd77abab6b2e71fa29ca4375d.png

Thank you!

Here are a few useful facts:
i. For x>0, sin(x) < x and therefore, sin(pi/(2n +1) < pi/(2n+1), n=1,2,3....
ii. n(3+ sin(n(pi)/4) > n for n=1,2,.....
Therefore, 0<=[sin(pi/(2n +1)]/[ n(3+ sin(n(pi)/4)] < pi/(n(2n+1)) < pi/n^2.
So, sum [sin(pi/(2n +1)]/[ n(3+ sin(n(pi)/4)] < sum pi/n^2
Since the series {pi/n^2} converges, so does {[sin(pi/(2n +1)]/[ n(3+ sin(n(pi)/4)]} .

oo
Σ sin(pi/(2n+1)) / (n(3 + sin(npi/4)))
n=1

Let's first note that the value of both sin(pi/(2n+1)) and sin(npi/4) are restricted between 1 and -1. Therefore, we can regard these values as not significant figures. Let's use the the limit comparison test to test for convergence. Let's let:

b_n = 1/n.

lim n-->oo [sin(pi/(2n+1)) / (n(3 + sin(npi/4)))] / (1/n)
lim n-->oo sin(pi/(2n+1)) / (3 + sin(npi/4)) = DNE.

Remember that these values are always going to alternate between two values. Therefore, the limit doesn't exist. The series diverges.

Hope this helped.