# If $a_n > 0$ and $\lim_{n \to \infty}{\sqrt{a_n}}>1$ show that $\lim_{n \to \infty}a_n$ exists.

If $a_n > 0$ and $\lim_{n \to \infty}{\sqrt{a_n}}>1$ show that $\lim_{n \to \infty}a_n$ exists.

I only managed to prove that there exists a sub-sequence which converges, but I couldn't prove the whole sequence does.

• Try by the definition of limit – user Apr 4 '18 at 8:54
• The $> 1$ condition isn't required. – Martín-Blas Pérez Pinilla Apr 4 '18 at 11:04
• @gimusi can you give me a bigger hint about using the definition? That's the way I did it for the sub-sequence, but could do it for $a_n$ – Jason Apr 5 '18 at 8:57
• @Jason I've added an answer with a trace for a proof by the definition. – user Apr 5 '18 at 9:41

## 2 Answers

By definition of limit assuming WLOG $\forall\epsilon \quad 0<\epsilon <1$

$$\exists n_0\quad \forall n\ge n_0 \quad |\sqrt{a_n}-L|<\epsilon\implies L-\epsilon<\sqrt{a_n}< L+\epsilon \\ \implies L^2-2\epsilon L+\epsilon^2< a_n<L^2+2\epsilon L+\epsilon^2 \implies -2\epsilon L+\epsilon^2< a_n-L^2<2\epsilon L+\epsilon^2\\\implies |a_n-L^2|< 2\epsilon L-\epsilon^2=\epsilon_1$$

thus $\forall\epsilon_1\quad 0<\epsilon_1 <2L-1$ we can choose $\epsilon \quad 0<\epsilon <1$ such that $2\epsilon L-\epsilon^2=\epsilon_1$ and

$$\exists n_0\quad \forall n\ge n_0 \implies|a_n-L^2|< \epsilon_1$$

then by definition

$$\lim_{n \to \infty}{a_n}=L^2$$

• Awesome! Thanks! – Jason Apr 5 '18 at 11:58
• @Jason You are welcome! Bye – user Apr 5 '18 at 11:59

$f(x)=x^2$ is continuous, so if $\lim_n\sqrt a_n$ exists and is strictly positive, so is $\lim_n(\sqrt a_n)$.