Test the convergence of a series To test the convergence of a series:
$$
\sum\left[\sqrt[3]{n^3+1}-n\right]
$$
My attempt:
Take out $n$ in common: $\displaystyle\sum\left[n\left(\sqrt[3]{1+\frac{1}{n^3}}-1\right)\right]$.
So this should be divergent.
But, the given answer says its convergent.

$$ \sqrt[3]{n^3+1}-n = \left({\sqrt[3]{n^3+1}-n)}\right) \frac{ \left({ (n^3 + 1)^{\frac 2 3} + (n^3 + 1)^{\frac 1 3}n  + n^{2 } }\right)}{\left({ (n^3 + 1)^{\frac 2 3} + (n^3 + 1)^{\frac 1 3}n  + n^{2 } }\right)} = \frac {1}{\left({ (n^3 + 1)^{\frac 2 3} + (n^3 + 1)^{\frac 1 3}n  + n^{2 } }\right)} \le \frac{1}{(n^3 + 1)^{\frac 2 3} } \le \frac{1}{(n^3)^{\frac 2 3}} = \frac 1 {n^2}$$
And we know that $\sum \frac {1}{n^2}$ converges and by the Comparison Test our series converges too. 

And a series will not diverge just because $n$ is a factor. What about $$ \sum n \cdot \frac{1}{n!} \;\;\; ?? $$
