Thread: Proof of a limit of an infinite sequence

1. Proof of a limit of an infinite sequence

Find and prove

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

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 $\displaystyle \frac{\sqrt{n+3}+ \sqrt{n}}{\sqrt{n+3}+ \sqrt{n}}$

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

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

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

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

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

and always greater than
$\displaystyle c_n=0$

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

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

and always greater than
$\displaystyle c_n=0$

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

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

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

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

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

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

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

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

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

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

Okay Perfect!