Divergence/convergence of a series how can I prove this series diverges/converges? I used all adequate criteria but nothing useful came out..any ideas ?
$$\sum\limits_{n=1}^\infty{\frac{{\sin (n)\cos (n^2 )}}{\sqrt n +\sqrt[3]n}}$$
 A: 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.
