1
$\begingroup$

So the theorem I am having trouble understanding is

If sequence $a_n$ converges and has the limit $L$, written $\lim_{n\rightarrow\infty}a_n=L$ if for every $\epsilon > 0$ there exists a positive integer $N$ such that $|a_n-L|<\epsilon$ whenever $n>N$

I am really not understanding how to implement this theorem. I am trying to use this theorem on the question

Determing whether $a_n$ converges or diverges, if it converges find its limit. $a_n=\sqrt{n+4}-\sqrt{n}$.

I know the limit is $0$ and it converges, but I am not understanding how to use that theorem. Thanks for all the help in advance.

$\endgroup$
3
  • $\begingroup$ Hint: Try to write $N$ in terms of $\epsilon$. $\endgroup$
    – BeaumontTaz
    Commented Sep 16, 2014 at 16:51
  • $\begingroup$ I'm trying, but I cannot come up with anything. How would I do that? @BeaumontTaz $\endgroup$
    – Kenshin
    Commented Sep 16, 2014 at 17:01
  • 1
    $\begingroup$ Another hint would be: $$\sqrt{n+4}-\sqrt{n}=\frac{4}{\sqrt{n+4}+\sqrt{n}}$$ If I give you the next step, the problem solves itself. Try to see if you can figure it out from here. $\endgroup$
    – BeaumontTaz
    Commented Sep 16, 2014 at 17:13

1 Answer 1

Reset to default
1
$\begingroup$

You want $\sqrt{n+4} - \sqrt{n} < \epsilon $. Well here's some useful stuff: $$\sqrt{n+4} - \sqrt{n} = \frac{4}{\sqrt{n+4} + \sqrt{n}} < \frac{4}{2\sqrt{n}}= \frac{2}{\sqrt{n}} $$

Now $\frac{2}{\sqrt{n}} < \epsilon$ if and only if $n > (\frac{2}{\epsilon})^2$.

Now let $\epsilon > 0$ be arbitrary. Let $N = \lceil(\frac{2}{\epsilon})^2\rceil $, then $\forall n > N$,$|a_n - 0| = \sqrt{n+4} - \sqrt{n} = \frac{4}{\sqrt{n+4} + \sqrt{n}} < \frac{4}{2\sqrt{n}}= \frac{2}{\sqrt{n}} < \epsilon $

Note $\lceil x \rceil$ is the ceiling function, it denotes the smallest integer that is greater than or equal to $x$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .