# 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 $\lim _\limits {n \to \infty}a_n=\infty$

real-analysissolution-verification

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

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*}