# Proving a sequence is monotone.

1. Nov 4, 2012

### peripatein

Hi,

I am unsuccessful at showing that the sequence √n + 1/n is an ascending monotone one, i.e. that a_n+1 > a_n for any n, greater than 2 let's say. I have proven that it is not bounded from above and is bounded from below. Any ideas, suggestions, please?

2. Nov 4, 2012

### SammyS

Staff Emeritus
This appears to be essentially the same as a previous thread you started:

3. Nov 4, 2012

### peripatein

Hi,
Except that no one there was able to help me. Perhaps someone could now.

4. Nov 4, 2012

### Zondrina

Consider your sequence as a function.

5. Nov 4, 2012

### peripatein

I am not allowed to. It has to be proven without any reference to functions. Seemingly merely by showing that a_n+1 > a_n for any n greater than 2. Yet I am unable to show that that inequality holds.

6. Nov 4, 2012

### Zondrina

Write out what a_(n+1) actually is.

7. Nov 4, 2012

### peripatein

sqrt(n+1) + 1/(n+1).

8. Nov 4, 2012

### Zondrina

Now what do you do to prove one sequence is larger than the other for all n≥2.

9. Nov 4, 2012

### peripatein

Induction? I have been unsuccessful at proving it via induction as well.

10. Nov 4, 2012

### Zondrina

If $a_{n+1} > a_n \Rightarrow a_{n+1} - a_n > 0$

11. Nov 4, 2012

### peripatein

And?

12. Nov 4, 2012

### Zondrina

Write out what it means.

13. Nov 4, 2012

### peripatein

sqrt(n+1) + 1/(n+1) - sqrt(n) - 1/n > 0

14. Nov 4, 2012

### Zondrina

You could clean it up a bit, but is that sequence greater than zero for every n?

If it helps, roughly around n = 1.2, your sequence geometrically will start to be above the line y=0.

Last edited: Nov 4, 2012
15. Nov 4, 2012

### peripatein

It surely is greater than zero but I am unable to demonstrate it algebrically, alas.

16. Nov 4, 2012

### Zondrina

For n=2 does the inequality hold?

17. Nov 4, 2012

### peripatein

It does.

18. Nov 4, 2012