# Convergence/Divergence and Monotonicity of a Sequence

1. Mar 20, 2013

### B18

1. The problem statement, all variables and given/known data
Show the following sequence to diverge, or converge. Determine if monotonic.
a sub n=n+(1/n)

3. The attempt at a solution
I understand that the sequence does diverge. I found this because the limit as n→∞ the limit is going to ∞ as well.
I found that the sequence is monotonic by showing that a sub n is less than a sub n+1. I tried to do the first derivative test for this and was confused when I had the sequence decreasing from 0 to 1 and increasing from 1 to ∞. Wouldn't this make the sequence not monotonic? Just need an explanation to this.

2. Mar 20, 2013

### LCKurtz

Surely you could show $a_n<a_{n+1}$ without resorting to calculus. But the sequence only has values on the integers 1,2,... so it would still be increasing.

3. Mar 20, 2013

### B18

So we are saying since sequences are only accountable for positive integers meaning the negative result on the first derivative test from 0 to 1 has no effect on monotocity.

4. Mar 20, 2013

### Jorriss

Why did you try the first derivative test on a sequence??

5. Mar 20, 2013

### B18

To determine if the sequence is increasing or decreasing to show it is monotonic

6. Mar 20, 2013

### LCKurtz

Yes. And I'm also saying that using the derivative test on $x+\frac 1 x$ is way overkill, not to mention the fact that it confused the issue for you.

7. Mar 20, 2013

### B18

Would I be correct in saying this sequence has a lower bound of 0 but no upper bound?

8. Mar 21, 2013

### LCKurtz

Is $a_n>0$ for all n? What do you think? Is that the greatest lower bound?
Is $a_n > n$? Again, what do you think?

9. Mar 21, 2013

### B18

Yes I believe that 0 is the greatest lower bound. And after plotting a couple of points on a graph of this sequence it appears this is no least upper bound.

10. Mar 21, 2013

### LCKurtz

The problem with "beliefs" is that they are sometimes wrong. And saying "it appears" is a long way from a proof. If this is a homework problem, you have a ways to go before handing it in.

11. Mar 21, 2013

### B18

Well because there is not an m that is ≥ a sub n this tells me there is no least upper bound. I know the greatest lower bound is 0 because the sequence never goes below 0 and therefore there is an M that is ≤ to all a sub n. according to my class notes this is as much evidence as I can provide. That we have learned thus far.

12. Mar 21, 2013

### LCKurtz

That argument shows 0 is a lower bound, but not that it is the greatest lower bound.

13. Mar 21, 2013

### B18

Do you have a quick explanation as to what makes a bound the least or greatest? Not seeing how 0 isn't the greatest lower bound when the sequence cannot go below that.

Last edited: Mar 21, 2013
14. Mar 21, 2013

### LCKurtz

What is the definition for x to be the greatest lower bound of a set S of numbers?