# Basic sequence related question

1. Jan 14, 2009

### duffman

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

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) }

2. Relevant equations

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

3. 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), couldnt acheive it...

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

Thanks,
Pierre-ALexandre.

2. Jan 14, 2009

### JG89

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

3. Jan 14, 2009

### duffman

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

4. Jan 14, 2009

### JG89

Double post, see below

5. Jan 14, 2009

### 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.

Last edited: Jan 14, 2009
6. Jan 14, 2009

### 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?

7. Jan 15, 2009

### HallsofIvy

Staff Emeritus
$$(-1)^n\frac{n+1}{n}$$
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.

8. Jan 15, 2009

### duffman

Well, that's much more simpler that way! I havent done calculus in 3 years so I kinda lost my intuition. Ugh! Thank you very much people!