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???
 A: Suppose the sequence converges to $c$. Can you see how your definition of the sequence implies that $c>0$?
Then for any $\epsilon>0$, the sequence eventually exceeds $c-\epsilon$.
If $c-\epsilon < x_n \le c$ then $x_n+\dfrac 1 {x_n^2}>{?}+\dfrac 1 {?}$. Can you fill in the blanks and see how this leads to a contradiction?
Showing the sequence increases without bound
Every increasing sequence of real numbers that is bounded above converges. As $(x_n)$ is an increasing sequence of real numbers that does not converge, it must be unbounded above.
A: 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.
A: Since $x_{n+1}> x_n$ here, it implies that the limit of $x_n$ is either a finite number or goes to infinity.
The first case is impossible, as 
$$\lim_{n\rightarrow\infty} x_n =l \in (1, \infty) \Rightarrow l+\frac{1}{l^2}=l \Rightarrow 1=0\text{(contradiction)}$$ Thus, $\lim_{n\rightarrow\infty} x_n = \infty$.

Added:(Proving that the sequence isn't bounded above)Since the sequence $x_n$ is increasing, then if it is convergent, it should have upper boundary and its limit is equal to its least upper bound (This means that if $l$ is an upper bound for $x_n$, for any other upper bound $L$ of $x_n$, we have $l \le L$). Suppose the limit of $x_n$ is $l$, which implies $x_n \lt l$. Also,by the definiton of convergent sequences, we have:
$$\forall \epsilon\gt0, \exists N\gt0,\text{such that, }n\gt N, \left|x_n-l\right|\lt \epsilon$$

hence $l-\epsilon\lt x_n\lt l$. Let $\epsilon = \frac{1}{l^2}$,then by the given formula above:
$$x_{n+1} = x_n + \frac{1}{x_n^2} \gt l-\epsilon + \frac{1}{l^2} = 1-\frac{1}{l^2}+\frac{1}{l^2} = l$$
which contradict to the fact that $x_{n+1} \lt l$. Hence, the upper bound do not exist,$x_n$ is divergent.
A: $\newcommand{\+}{^{\dagger}}%
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$\large\tt Hint:$
$$
x_{n} = x_{n - 1} + {1 \over x_{n - 1}^{2}}
=
x_{n - 2} + {1 \over x_{n - 2}^{2}} + {1 \over x_{n - 1}^{2}}
=
\cdots
=
x_{1} + {1 \over x_{1}^{2}} + {1 \over x_{2}^{2}} + \cdots + {1 \over x_{n - 1}^{2}}
$$
$$
x_{n} = x_{1} + \sum_{k = 1}^{n - 1}{1 \over x_{k}^{2}}\,\qquad n \geq 2
$$
If $\lim_{n \to \infty}x_{n}$ is finite, the 'generic serie term' should go to zero when $n \to \infty$: $\quad\lim_{k \to \infty}\pars{1/x_{k}^{2}} = 0$ which contradicts the initial statement. Then, $\lim_{n \to \infty}x_{n} = \infty$. 
