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This is rather unpleasant, but totally do-able.

Convergent:

First, rewrite the numerator using the product-to-sum formula, as follows:

$$\sum\limits_{n=1}^\infty{\frac{{\sin (n)\cos (n^2 )}}{\sqrt n +\sqrt[3]n}} = \frac{1}{2}\sum\limits_{n=1}^\infty{\frac{{\sin (n(n+1))-\sin (n(n-1) )}}{\sqrt n +\sqrt[3]n}}$$

Notice how the numerator appears to be telescoping a bit. If you write out the first few terms, the pattern is pretty easy to pick out.

$$\frac{1}{2}\sum\limits_{n=1}^\infty{\frac{{\sin (n(n+1))-\sin (n(n-1) )}}{\sqrt n +\sqrt[3]n}} $$

$$= \frac{1}{2}\sum\limits_{n=1}^\infty{\sin (n(n+1))\bigg(\frac{1}{\sqrt n + \sqrt[3] n} - \frac{1}{\sqrt {n+1} + \sqrt[3] {n+1}}\bigg)}$$

Next, since:

$$\frac{1}{2}\sum\limits_{n=1}^\infty{\bigg|\sin (n(n+1))\bigg(\frac{1}{\sqrt n + \sqrt[3] n} - \frac{1}{\sqrt {n+1} + \sqrt[3] {n+1}}\bigg)\bigg|}$$

$$ \leq \frac{1}{2} \sum\limits_{n=1}^\infty{\bigg(\frac{1}{\sqrt n + \sqrt[3] n} - \frac{1}{\sqrt {n+1} + \sqrt[3] {n+1}}\bigg)} \leq \infty$$ (by the integral test)

...we have that the series is absolutely convergent.

Edit: Clarity.