<s>[INST] Test the convergence of the series $\sum [(n^3+1)^{\frac{1}{3}}- n]$ 
Test the convergence of the series $\sum [(n^3+1)^{\frac{1}{3}}- n]$

I have no idea which test to use here because of this complicated expression.
 [/INST] $n^{th}$ term of the series is 
= $ (n^3+1)^{\frac{1}{3}}- n$
= $ n(1+\frac{1}{n^3})^{\frac{1}{3}}- n$
= $ n(1+\frac{1}{3n^3} + other terms)- n$ . . . . . .  . .(other terms goes to 0 as n approaches $\infty$)
= $ \frac{1}{3n^2}$ + other terms
Now as $n\to\infty$ ,our $n^{th}$ term can be shown with following inequality
= $ \frac{1}{3n^2}$+other terms$ \le \frac{1}{n^2} < \frac{1}{n}$
Since $n^{th}$ term of series is less than $\frac{1}{n}$, the series is convergent
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