Question

Determine if the series converges or diverges: an=\[\sqrt{n}-\sqrt{n^2-1}\]

Answer 1

1) Do the successive terms approach zero (0)? If "no", then the answer is "no"

Given that 1) is true...

2) Do they alternate in sign? Done! Yes. It converges.

Given that 1) is true and 2) is false... Well, it has a chance. It's time to pull out all your tests and figure it out.

In this case, you may wish to "rationalize" the numerator...

Hold on, have you written the problem statement correctly? Are you SURE it's not just \(n\), rather than \(\sqrt{n}\)?

Answer 2

No, I believe this diverges to - infinity.
If you take the limit you get \[\lim_{n \rightarrow \infty} \sqrt{n}-\sqrt{n^2-1} \]
You can clearly see that it will diverge to -infinity, but I am not sure how to show the more detailed work to the limit. Could I divide both parts of the limity by sqrt(n)?

Answer 3

Prove successive terms fail to decrease in magnitude. That's one reason we invented the Ratio Test. It might work.

Answer 4

@tkhunny we haven't talked about the ratio test in class, but I will take a look at it and ask the TA if we can use it. Thanks!

Answer 5

If anyone knows how to do it without the ratio test and could post it, that would be awesome! :D

Answer 6

You can also just subtract two successive terms. If you can prove the expression is positive, we're done.

\(\left(\sqrt{n+1}-\sqrt{(n+1)^{2}-1}\right)-(\sqrt{n}-\sqrt{n^{2}-1})\)

\(\sqrt{n+1}\left(1 + \sqrt{n-1}\right) - \sqrt{n}\left(1+\sqrt{n+2}\right) > \sqrt{n}\left(1 + \sqrt{n-1}\right) - \sqrt{n}\left(1+\sqrt{n+2}\right)\)

\(\sqrt{n}\left(\sqrt{n-1}+\sqrt{n+2}\right)\) -- Now THAT looks positive.

Answer 7

The ratio test is not helpful on this one. There's a funny and distressing part about the Ratio Test. Sometimes it fails! This is the case on this one.

Answer 8

Yup so that means the series diverges, as a(n+1)>an