Find Limit of Sequence: \sqrt{n} - \sqrt{n^2 - 1}

[G01]

I have to find out where this sequence converges or if it converges a all:

$$a_n = \sqrt{n} - \sqrt{n^2 - 1}$$

Now, I can't seem to find a good method to solve this. Would my best bet be to use l'hospital's rule to find the limit of the equivalent function or should I try the squeeze theorem. Thats my question. Thanks for the help.

 

[durt]

It looks like it diverges, so show that it is bounded by a divergent sequence.

 

[G01]

ok how the this sound as a complete solution?

An tends seems to tend toward negative infinity and diverge. I proove this by showing that a function which is always greater than it also tends toward negative infinity. I take the greater function to be the squareroot of n.([tex]f(x) = \sqrt{x}$$} This function tends toward negative infinity, so the corressponding sequence, and since this sequence is less than the sequence given, the sequence given will also tend toward negative infinity.<br /> <br /> Of course in my solution i'll use more mathmatical notation, but is that the correct reasoning?$$[/tex]

 

[durt]

Yeah that's right. Except I think you mean the negative square root: $$f(x)=-\sqrt{x}$$. And make sure you have in there: $$f(n) > a_n$$ for all $$n > a$$ (where you find a).