# Proof of a limit of an infinite sequence

• Mar 15th 2013, 04:21 PM
Kanwar245
Proof of a limit of an infinite sequence
Find and prove

$\lim_{n\rightarrow\infty}(\sqrt{n+3} - \sqrt{n})$
• Mar 15th 2013, 04:21 PM
Kanwar245
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
• Mar 15th 2013, 04:24 PM
HallsofIvy
Re: Proof of a limit of an infinite sequence
That sounds good- multiply by $\frac{\sqrt{n+3}+ \sqrt{n}}{\sqrt{n+3}+ \sqrt{n}}$
• Mar 15th 2013, 04:33 PM
Kanwar245
Re: Proof of a limit of an infinite sequence
yes and it becomes
$\frac{3}{(\sqrt{n+3}+\sqrt{n})}$
• Mar 15th 2013, 05:06 PM
Shakarri
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.
• Mar 15th 2013, 05:32 PM
Prove It
Re: Proof of a limit of an infinite sequence
Quote:

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.
• Mar 15th 2013, 05:44 PM
Shakarri
Re: Proof of a limit of an infinite sequence
Oh yes you could do it that way (Clapping)
• Mar 15th 2013, 06:36 PM
Kanwar245
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...
• Mar 15th 2013, 06:53 PM
Prove It
Re: Proof of a limit of an infinite sequence
Quote:

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).
• Mar 15th 2013, 07:18 PM
Kanwar245
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 ${\epsilon}^2$ ...
• Mar 15th 2013, 08:59 PM
Prove It
Re: Proof of a limit of an infinite sequence
Quote:

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?
• Mar 16th 2013, 07:03 AM
Kanwar245
Re: Proof of a limit of an infinite sequence
Quote:

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 $N = \frac{9}{4\epsilon^2}$?
• Mar 16th 2013, 03:05 PM
Prove It
Re: Proof of a limit of an infinite sequence
Quote:

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 :)
• Mar 16th 2013, 03:06 PM
Kanwar245
Re: Proof of a limit of an infinite sequence
Okay Perfect!