Let $a_n$ be a sequence such that $a_1 >0$ and for all $n$ $a_{n+1}=a_n+ \frac{1}{a_n^3}$ then $\lim _\limits {n \to \infty}a_n=\infty$ 
Let $a_n$ be a sequence such that $a_1 >0$ and for all $n$ $a_{n+1}=a_n+ \frac{1}{a_n^3}$ then  $\lim _\limits {n \to \infty}a_n=\infty$


I am not sure how to formally prove this , the book says the statement is true.
I just tried to gather some information from what is given I do not know if it is mostly needed for example according to $a_{n+1}=a_n+ \frac{1}{a_n^3}$ we get that $a_{n+1}-a_n=\frac{1}{a_n^3}$ so the sequence is increasing
also we know that the limit of $a_n$ is equal to the limit of a moved sequence (sorry if this is the incorrect word) so $\lim _\limits {n \to \infty}a_n=\lim _\limits {n \to \infty}a_{n+1}$
and the only way to get this it is if I can get $\lim _\limits {n \to \infty}\frac {1}{a_n^3}=0$ and that is when $\lim _\limits {n \to \infty}a_n= \infty$
This was my approach , according to all that we get that $\lim _\limits {n \to \infty}a_{n+1}= \lim _\limits {n \to \infty}a_n+ \frac{1}{a_n^3}$ since $\lim _\limits {n \to \infty}a_n= \infty$
the result will be $\lim _\limits {n \to \infty}a_{n+1}= \lim _\limits {n \to \infty}a_n=\infty$
Is my way correct? is there a different approach?
Thank you for any tips and help
 A: For every $0\lt a_1\lt 1,a_2\ge 1$.  Since $a_1\gt 0$ and $a_{n+1}=a_n+\frac 1{a_n^3}\implies a_{n+1}\ge a_n$, we have that $a_n$ is monotone increasing.
Choose an arbitrary $M\gt a_1$.  Then there exists a $q\ge 1$ such that for every $r=1,2,\dots,q$ we have $a_r\lt M$ which in turn means $\frac 1{a_r^3}\ge \frac 1{M^3}$.  Taking $q\ge M^4$ we get $\sum_{r=1}^{M^4}\frac 1{a_r^3}\ge\sum_{r=1}^{M^4}\frac 1{M^3}=M$.  Then $a_{q}\ge M$, showing that $q$ was chosen incorrectly and there is some $q_1\lt q$ such that $a_{q_1}\lt M\le a_{q_1+1}$ and in particular $a_{q_1+2}\ge M+\frac 1{a_{q_1+1}^3}\gt M$.  But $M$ was chosen arbitrarily, so there is no upper bound on the sequence $a_n$.
A: Squaring both sides,
\begin{align*}
a_{n+1}^2 = a_n^2 + 2\frac{a_n}{a_n^3} + \frac{1}{a_n^6} > a_n^2 + 2\frac{1}{a_n^2}
\end{align*}
and squaring again,
\begin{align*}
a_{n+1}^4 > a_n^4 + 4 + \frac{4}{a_n^4} > a_n^4 + 4
\end{align*}
and so
\begin{align*}
a_{n+1}^4 = a_1^4 + \sum_{k=1}^{n}(a_{k+1}^4 - a_k^4) > 4n + a_1^4 \rightarrow \infty
\end{align*}
