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.
 A: $$\sqrt[3]{n^3+1}-n=\frac{1}{\sqrt[3]{n^3+1}^2+n\sqrt[3]{n^3+1}+n^2}\le\frac{1}{3n^2}$$
A: We have that $$\sqrt[3]{1+\frac{1}{n^3}} \sim_\infty 1+\frac{1}{3n^3}$$ by Taylor series. 
So $$n\left(\sqrt[3]{1+\frac{1}{n^3}}-1\right)\sim_\infty\frac{1}{3n^2}.$$
Do you see why the series is convergent now? :)
A: $$ \sqrt[3]{n^3+1}-n = \left({\sqrt[3]{n^3+1}-n)}\right) \frac{ \left({ (n^3 + 1)^{\frac 2 3} + (n^3 + 1)^{\frac 1 3}n  + n^{2 } }\right)}{\left({ (n^3 + 1)^{\frac 2 3} + (n^3 + 1)^{\frac 1 3}n  + n^{2 } }\right)} = \frac {1}{\left({ (n^3 + 1)^{\frac 2 3} + (n^3 + 1)^{\frac 1 3}n  + n^{2 } }\right)} \le \frac{1}{(n^3 + 1)^{\frac 2 3} } \le \frac{1}{(n^3)^{\frac 2 3}} = \frac 1 {n^2}$$
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 $$ \sum n \cdot \frac{1}{n!} \;\;\; ?? $$
A: This is quite similar to the other answers, but slightly different derivation of the estimate:
$$n^3+1 \le n^3+3 +\frac3{n^3}+\frac1{n^6}=\left(n+\frac1{n^2}\right)^3$$
$$\sqrt[3]{n^3+1} \le n+\frac1{n^2}$$
$$\sqrt[3]{n^3+1} -n \le \frac1{n^2}$$
Now we can use comparison test.
If you prefer to have  a tighter estimate, you can use $\left(n+\frac1{3n^2}\right)^3$ instead.
