Does the Sequence {(-1)^n*(n+1/n)} Converge or Diverge?

[duffman]

Homework Statement

Here is the problem as deployed in the exercise book:
The point is to say if the sequence converges or diverges
an= { (-1)^n*(n+1/n) }

Homework Equations

Bluntly enough: what's the key to finding the result?

The Attempt at a Solution

I tried applying l'Hopital's rule but got lost into oblivion. I tried finding a smaller and bigger sequences (sandwich propriety), couldn't achieve it...

The answer guide says the sequence diverges. Why, and how do you get there?

Thanks,
Pierre-ALexandre.

 

[JG89]

Do you mean (-1)^n * (\frac{n+1}{n}) or (-1)^\frac{n(n+1)}n}

 

[duffman]

The first one is the correct sequence. Thanks for cleaning it up :) (sorry, noob alert :/)

 

[JG89]

Double post, see below

 

[JG89]

For even n we have a_n = 1*\frac{n+1}{n} = \frac{n+1}{n}. For odd n we have a_n = -1*\frac{n+1}{n} = \frac{-(n+1)}{n}.

Remember that a sequence converges if and only if there is a positive integer N such that when integers n,m > N, we have |a_n - a_m| < \epsilon for all epsilon greater than 0. That should help you out :).

 

[Dick]

If your sequence is ((-1)^n)*((n+1)/n) then that's ((-1)^n)*(1+1/n). The limit of the second factor is 1. So for large n your sequence looks like +1,-1,+1,-1,... Is that convergent?

 

[HallsofIvy]

(-1)^n\frac{n+1}{n}
Divide both numerator and denominator by n:
(-1)^n\frac{1+\frac{1}{n}}{1}= (-1)^n\left(1+ \frac{1}{n}\right)
If this were just
\left(1+ \frac{1}{n}\right)
then it should be easy to see what the limit is as n goes to infinity. But with the (-1)^n think about the subsequences for n even or odd.

 

[duffman]

Well, that's much more simpler that way! I haven't done calculus in 3 years so I kinda lost my intuition. Ugh! Thank you very much people!