# Converging Sequence

1. May 9, 2007

### Lucretius

1. The problem statement, all variables and given/known data

Determine whether each of the sequences converges or diverges. If it converges, find its limit. Explain with sufficient details each claim.

2. Relevant equations
(k) $$a_n=\sqrt{n^2+1}-n$$

The Squeeze Theorem (If $$b_n<a_n<c_n$$, and both $$\displaystyle\lim_{x\rightarrow\infty}a_n, c_n = L$$, then $$\displaystyle\lim_{x\rightarrow\infty}b_n=L$$)

3. The attempt at a solution

I had the choice of deciding between the squeeze theorem and L'Hopital. L'Hopital seemed useless because I don't see a way to get this into the form of a quotient. I know the sequence is convergent to 0. Figured it would be best to try the Squeeze theorem, as it's really my only other option.

I know that $$0 \leq \sqrt{n^2+1}-n$$, but after struggling for a while, I still have no idea how to get a larger limit on the other side equal zero (without arbitrarily choosing one, 1/x for instance works, but there's no logical way to get from my function to 1/x...) I know that $$n+1>\sqrt{n^2+1}$$, but the result is n+1-n=1, and the limit of 1 is 1, period. Likewise, I can get rid of the 1, but I'll have n left over, and the limit of n is infinity! Got any good tips on getting the other side of the squeeze theorem inequality?

2. May 9, 2007

### quasar987

Multiply by $$\frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}$$ and watch the magic.

3. May 9, 2007

### AKG

You stated the Squeeze theorem incorrectly, but I suppose you know the correct statement. I don't know what you mean by there being "no logical way" to get 1/n, but if 1/n works, then it works, so what's the problem? 1/n converges to 0 as n goes to infinity, if you can prove that $\sqrt{n^2+1}-n\leq n^{-1}$ for all n, then you're done. There's an easier way in my opinion, but do it your way first, and once you have your own answer I can give you mine.

4. May 10, 2007

### Lucretius

Ah thx quasar; I shoulda thought of rationalization. I need to remember that everytime I see a square root...

When I meant "logical way" AKG, I meant that, just by looking at the equation, we can't conclude that 1/x is bigger always. I had to look at a calculator. I have to justify my choice for the boundaries of the squeeze theorem. However, quasar pointed out I could just rationalize the function and end up finding out that 1/n is bigger.

5. May 10, 2007

### AKG

No, but you should be able to prove it.