# Help finding where a sequence converges

1. Sep 30, 2012

### rocapp

Hi all,

I have a general question along with a specific one. Could anyone give me an intuitive explanation of finding limits of convergent sequences? I have a test and just do not understand how to consistently find the answer to these problems. I can understand intuitively obvious ones, but for ones like the example below, I am stumped.

Here's my specific question-

This sequence converges at (1/2)sqrt(2). How would one find this?

(n^2)/sqrt((2n^4)+1)

Thanks!

Rob

2. Sep 30, 2012

### Dick

Pull a factor of n^2 out the denominator and cancel it with the factor of n^2 in the numerator. What's left?

3. Sep 30, 2012

### rocapp

You are a genius. Haha. Thanks!

4. Oct 1, 2012

### rocapp

Actually, I still do not understand. Removing a term of n^2 will give 1/(2n^2+1)^(1/2).

How would you get sqrt(2)/2 from this?

5. Oct 1, 2012

### rocapp

Would the answer not be 1/2?

My book may have an error.

Sorry for the double post.

6. Oct 1, 2012

### Dick

Your algebra is wrong. To get a factor of $n^2$ out of $\sqrt{2n^4+1}$ first write it as $\sqrt{n^4(2+\frac{1}{n^4})}$. Now what??

7. Oct 1, 2012

### rocapp

The n^2 on top and bottom cancel. I'm left with 1/sqrt(2+(1/n^4)). Still not sure what to do next.

8. Oct 1, 2012

### Dick

Now take the limit n->infinity. What happens to the 1/n^4?

9. Oct 1, 2012

### rocapp

It goes to zero, and I'm left with 1/sqrt(2). Ohhhh, and that is equivalent to sqrt(2)/2, yes?

10. Oct 1, 2012

### Dick

Yes it is!

11. Oct 1, 2012

### rocapp

HaHA! Thanks a bunch! I have a test later and will definitely utilize my newfound limit skills.