Is $\sum_{n=1}^\infty \frac{\sin(2n)}{1+\cos^4(n)}$ convergent?

As I was just checking this 'child prodigy' out on Youtube, I stumbled upon this video, in which Glenn Beck asks the kid to do the following proof:

Further on, the kid starts sketching a proof (without a shadow of a doubt regarding the accuracy of his solution ) including the Integral Test. I don't know much about improper integrals since I just finished highschool, but this integral approach seemed, intuitively, pretty inaccurate to me since this is not a strictly decreasing function. Then I found this out from Wikipedia ! Conditions for the Integral Test.

Furthermore, the blunt assessment that the series are convergent seems dubious as well..Upon a few computations of my own(mostly partial sums) , I tend to believe that the series are, in fact, divergent .

Can anyone suggest a rigurous take on this problem (easy as it may seem to some amongst you) ?

Yes the series is indeed divergent. A necessary condition for the series to be convergent is, that the sequence $$\frac{\sin(2n)}{1+\cos^4(n)}$$ tends to zero as $n\to\infty$. This, however, is not the case as $$\left|\frac{\sin(2n)}{1+\cos^4(n)} \right|\geq \left|\frac{\sin(2n)}{2}\right|$$ and $\sin(2n)$ obviously does not tend to $0$.

\begin{align} \sin(2n) &= 2\sin(n)\cos(n) = -\frac{\mathrm i}{2}\sum_{n=0}^\infty\frac{(2\mathrm in)^n - (-2\mathrm i n)^n}{n!} \\ &= -\frac{\mathrm i}2 \sum_{n=0}^\infty \frac{n^n(2^n(\mathrm i^n - \mathrm i^{3n}))}{n!} \\ &= -\frac12 \sum_{n=1}^\infty \frac{(2\mathrm in)^n(1 - (-1)^n)}{n!}. \end{align}

Noting $$\lim_{n\to\infty}\left|\frac{(2n)^n\mathrm i^{n+3}}{n!}\right| = \infty,$$ and that the radius of curvature is $\limsup_{n\to\infty}{(n!/(2n)^n)^{1/n}} = \lim{n\to\infty}(2n)^{-1} = 0,$ and that $$1 \le (1+\cos^4 n) \le 2,$$ it is evident that both the real and imaginary parts of the series $$\sum_{n=1}^\infty\frac{\sin 2n}{1+\cos^4 n}$$ diverge.

Similarly, the same is true of both real and imaginary parts of denominator. Hence the series neither converges nor diverges.

Accomplished by expressing $\sin(n)\cos(n)$ exponentially and implementing the exponential serial expansion. Similarly the denominator.