# $(x_n)$ is a sequence of real numbers. Given $x_1=1$ and $x_{n+1}=x_n +\frac{1}{{x_n}^2}$. Show that the sequence is unbounded

I know there is an answer to this question. But I have a different way to prove this. Can someone help me to check if my proof is correct? Thanks.

Proof: Suppose the sequence $x_1,x_2,x_3,...$ of real numbers given by $x_1=1$ and $x_{n+1} = x_n +\frac{1}{x_n^2}$ for each $n=1,2,3,...$ is bounded.

Then $$\lim_{n\to \infty} x_n=x$$ for some $x\in \mathbb{R}$.

Hence $$\lim_{n\to \infty} x_{n+1}=\lim_{n\to \infty} (x_n +\frac{1}{x_n^2})=x+\frac{1}{x^2} \not = x=\lim_{n\to \infty} x_n.$$

So the supposition is false, and the sequence is unbounded.

• Given that the sequence is bounded, you need to say one more thing before you can conclude it converges... – Wojowu Oct 8 '17 at 20:58
• Pedagogically I recommend you to explicitly mention that $x \neq 0$. Other than that, your solution is fine. But I see no difference between your solution and the other solutions in the link. – Sangchul Lee Oct 8 '17 at 20:58
• You should have $\frac{1}{x^2}$ not $\frac{1}{x}$ just left of $\neq$. – Rob Arthan Oct 8 '17 at 21:01

## 1 Answer

$(x_n)$ is an increasing sequence.

if it is bounded above, it will converge and $x_{n+1}-x_n$ will go to zero.

but $$x_{n+1}-x_n=\frac {1}{x_n^2}$$

cannot go to zero. thus $(x_n)$ is unbounded above.

• Elegant argument! – Michael Lee Oct 8 '17 at 21:05