# Thread: Proof of a limit of an infinite sequence

1. ## Proof of a limit of an infinite sequence

Find and prove

$\lim_{n\rightarrow\infty}(\sqrt{n+3} - \sqrt{n})$

2. ## Re: Proof of a limit of an infinite sequence

I'm thinking of rationalizing it first, also I can see that the limit approaches 0

3. ## Re: Proof of a limit of an infinite sequence

That sounds good- multiply by $\frac{\sqrt{n+3}+ \sqrt{n}}{\sqrt{n+3}+ \sqrt{n}}$

4. ## Re: Proof of a limit of an infinite sequence

yes and it becomes
$\frac{3}{(\sqrt{n+3}+\sqrt{n})}$

5. ## Re: Proof of a limit of an infinite sequence

From
$a_n=\frac{3}{\sqrt{n+3}+\sqrt{n}}$

You can show that this is always less than
$b_n=\frac{3}{\sqrt{n}+\sqrt{n}}$

and always greater than
$c_n=0$

Using the sandwich theorem show that an converges to zero.

6. ## Re: Proof of a limit of an infinite sequence

Originally Posted by Shakarri
From
$a_n=\frac{3}{\sqrt{n+3}+\sqrt{n}}$

You can show that this is always less than
$b_n=\frac{3}{\sqrt{n}+\sqrt{n}}$

and always greater than
$c_n=0$

Using the sandwich theorem show that an converges to zero.
Isn't the sandwich theorem overkill here? The denominator clearly goes to infinity, so the whole fraction goes to 0.

7. ## Re: Proof of a limit of an infinite sequence

Oh yes you could do it that way

8. ## Re: Proof of a limit of an infinite sequence

right but when proving, for all epsilon greater than 0, what N would I choose so that $|\frac{3}{2\sqrt{n}}| < \epsilon$
If i try to square it, epsilon gets squared too...

9. ## Re: Proof of a limit of an infinite sequence

Originally Posted by Kanwar245
right but when proving, for all epsilon greater than 0, what N would I choose so that |3/2sqrt(n)| < epsilon
If i try to square it, epsilon gets squared too...
Why bother squaring anything? Since this is a sequence, we know n is positive, and so the stuff inside the absolute value has to be positive. This means its equal to its absolute value (or the absolute value sign can be neglected).

10. ## Re: Proof of a limit of an infinite sequence

Right so we have this:
$|\frac{3}{\sqrt{n} + \sqrt{n}}| < \epsilon$
$\Rightarrow |\frac{3}{2\sqrt{n}}| < \epsilon$
$\Rightarrow \frac{3}{2\sqrt{n}} < \epsilon$
Now from this point on, if I try to isolate n, at some point I will have to square $\sqrt{n}$ which will yield ${\epsilon}^2$ ...

11. ## Re: Proof of a limit of an infinite sequence

Originally Posted by Kanwar245
Right so we have this:
$|\frac{3}{\sqrt{n} + \sqrt{n}}| < \epsilon$
$\Rightarrow |\frac{3}{2\sqrt{n}}| < \epsilon$
$\Rightarrow \frac{3}{2\sqrt{n}} < \epsilon$
Now from this point on, if I try to isolate n, at some point I will have to square $\sqrt{n}$ which will yield ${\epsilon}^2$ ...
What's wrong with that?

12. ## Re: Proof of a limit of an infinite sequence

Originally Posted by Prove It
What's wrong with that?
Oh so we can do this?
$\sqrt{n} > \frac{3}{2\epsilon}$

$n > \frac{9}{4\epsilon^2}$

So when proving, I can say
Given $\epsilon > 0$
Choose $N = \frac{9}{4\epsilon^2}$?

13. ## Re: Proof of a limit of an infinite sequence

Originally Posted by Kanwar245
Oh so we can do this?
$\sqrt{n} > \frac{3}{2\epsilon}$

$n > \frac{9}{4\epsilon^2}$

So when proving, I can say
Given $\epsilon > 0$
Choose $N = \frac{9}{4\epsilon^2}$?
You certainly can

14. ## Re: Proof of a limit of an infinite sequence

Okay Perfect!