How do you evaluate a limit n->inf. if there's a (-1)^n term?

[pylauzier]

Homework Statement

I decided to work through my old calculus notes and I can't find any information about these limits. Basically I'm trying to evaluate lim n->∞ (-1)ⁿ *sqrt(n+1)/n.

Homework Equations

Not really applicable.

The Attempt at a Solution

Well, in this particular case it's easy to see that the limit will be zero if we divide the numerator and denominator by n, since the square root will tend towards zero. However, could this limit have been evaluated if the squared n was actually n²?

Thanks in advance!

 

[Dick]

You should be able to handle that one with the alternating series test. Now how exactly are you thinking about changing it?

 

[pylauzier]

What do you mean by "Now how exactly are you thinking about changing it?" ??

 

[Dick]

I mean this part. "However, could this limit have been evaluated if the squared n was actually n^2?". What do you mean by that?

 

[pylauzier]

Oh, basically I said that I managed to solve the first limit I posted since the numerator tends toward 0 as n -> ∞, so that one was obvious. However with n² instead of n under the square root, the limit would be +/- 1. If I recall correctly that's not a valid answer, so surely there's a way to evaluate further?

 

[Dick]

Oh, basically I said that I managed to solve the first limit I posted since the numerator tends toward 0 as n -> ∞, so that one was obvious. However with n² instead of n under the square root, the limit would be +/- 1. If I recall correctly that's not a valid answer, so surely there's a way to evaluate further?

If you are thinking about lim n->infinity (-1)^n*sqrt(n^2+1)/n then there is no limit. It alternates between points near +1 and -1, as you said.

 

[pylauzier]

If you are thinking about lim n->infinity (-1)^n*sqrt(n^2+1)/n then there is no limit. It alternates between points near +1 and -1, as you said.

Alright that answers my question, thanks!