1
$\begingroup$

I need help to make my proof absolutely correct.

We got $a_n = \sqrt{n} \cdot (\sqrt{n+3}-\sqrt{n})$. Proof that the sequence diverges.

Proof:

$\lim\limits_{n \rightarrow \infty} \sqrt{n} = \infty $

$ \sqrt{n+3} > \sqrt{n} \Rightarrow (\sqrt{n+3} - \sqrt{n}) > 0 \Rightarrow \lim\limits_{n \rightarrow \infty} (\sqrt{n+3} - \sqrt{n}) = \infty $

$\Rightarrow \lim\limits_{n \rightarrow \infty} \sqrt{n} \cdot (\sqrt{n+3}-\sqrt{n}) = \infty $

I do not know if my proof is perfectly correct. Please do not tell me that there are complete different ways to solve this because i am not allowed to use higher technics in my exam. So please help me to improve my solution.

  • 0
    The fact that $\left( \sqrt{n+3}-\sqrt{n} \right) > 0$ doesn't imply that it's limit is $\infty$. For instance, consider $\frac{1}{n}$ which is always greater than 0, but its limit as $n \to \infty$ is $0$. What other techniques do you have at your disposal? – 2017-01-20
  • 0
    Would it be okay if I change that to: $(\sqrt{n+3}-\sqrt{n})$ is unlimited $\Rightarrow (\sqrt{n+3}-\sqrt{n})$ diverges – 2017-01-20
  • 0
    @Arjihad: what you have is actually $\sqrt{n+3}-\sqrt{n} \rightarrow 0$. Take a look in my answer. – 2017-01-20

2 Answers 2

1

Hint

$$\sqrt{n}\cdot (\sqrt{n+3}-\sqrt{n})=\frac{\sqrt{n}\cdot(\sqrt{n+3}-\sqrt{n})(\sqrt{n+3}+\sqrt{n})}{(\sqrt{n+3}+\sqrt{n})}=\frac{3\sqrt{n}}{(\sqrt{n+3}+\sqrt{n})}=\frac{3}{\sqrt{1+\frac{3}{n}}+1}$$

Fot the last equality just divide both terms by $\sqrt{n}$.

Now you can do $n \rightarrow \infty$.

Can you finish?

  • 0
    Thank you very much! I see that $\dfrac{3}{ \sqrt{ 1+\frac{3}{n} }+1} $ converges to$ \dfrac{3}{2}$. – 2017-01-20
  • 0
    You are very welcome. Yes, the limit is $3/2$. – 2017-01-20
  • 0
    Done. Thanks for your help! – 2017-01-20
  • 0
    @Arjihad, You are very welcome! – 2017-01-20
0

Please note that while it is true that

$$\sqrt{n+3} - \sqrt{n} > 0$$

it is false that

$$\lim_{x\to \infty} \sqrt{n+3} - \sqrt{n} = \infty \tag{1}$$

given that we actually have

$$\lim_{x\to \infty} \sqrt{n+3} - \sqrt{n} = 0 \tag{2}$$

So the limit you want to find is actually a limit of the form $0 \cdot \infty$ because $\sqrt{n} \to \infty$ when $n \to \infty$.

Therefore we must do something else to find the limit. A suggested approach would be to multiply and divide by $\sqrt{n+3} + \sqrt{n}$ giving

$$\lim_{n \to \infty} \sqrt{n}\cdot \frac{(\sqrt{n+3}-\sqrt{n})(\sqrt{n+3}+\sqrt{n})}{\sqrt{n+3}+\sqrt{n}}$$

You should be able to simplify the numerator to get

$$\lim_{n \to \infty} \frac{3\sqrt{n}}{\sqrt{n+3}+\sqrt{n}}$$

Can you solve this new limit? First, convince yourself that when $n$ goes to infinity, $\sqrt{n+3} \approx \sqrt{n}$