Limit of $x_1=1, x_{n+1}=x_n+\frac1{x_n^2}$ 
Given that $x_1 = 1$ and $x_{n+1}=x_{n}+ 1/x_{n}^{2}$. Find the limit of the sequence.

Let $ c $ be the limit of the sequence, then $ c=c+\frac{1}{c^2} $, that means $ \frac{1}{c^2}=0 $. That can't be like that.What is wrong???

Notice $$x_{n+1}^3 = x_n^3 (1 + \frac{1}{x_n^3})^3 = x_n^3 + 3 + \frac{3}{x_n^3} + \frac{1}{x_n^6}$$
We have $x_{n+1}^3 - x_n^3 > 3$ for all $n$ and hence we can bound $x_n$ from below
$$x_n^3 \ge x_1^3 + 3(n-1) = 3n-2 \quad\implies\quad x_n \ge \sqrt[3]{3n-2}$$
As a result, the sequence $x_n$ diverges.
