# Monotonic increasing or monotonic decreasing

1. Feb 22, 2010

### Ki-nana18

1. The problem statement, all variables and given/known data
1. Determine whether the sequence {an} = n+(1/n) is monotonic increasing or monotonic decreasing.

2. Relevant equations

3. The attempt at a solution
I plugged in some digits and got this
a1=2
a2=5/2=2.5
a3=10/3=3.3333333
a4=17/4=4.25
a5=26/5=5.2
I drew the coclusion that it is monotonic increasing. Is that right?

2. Feb 22, 2010

### Staff: Mentor

Re: Sequences

You need to show that an+1 >= an for all n >= some number M. You can't just use the values of a few elements of the sequence.

3. Feb 22, 2010

### Ki-nana18

Re: Sequences

What is M? How do I find it?

4. Feb 22, 2010

### VeeEight

Re: Sequences

Try calculating an+1 - an

5. Feb 22, 2010

### Staff: Mentor

Re: Sequences

You get to say what it is.

6. Feb 22, 2010

### Ki-nana18

Re: Sequences

n+1+(1/n+1)>= n+(1/n) @n=3
4.25>3.33333

Since an+1 is greater that an the sequence is monotonic increasing.

7. Feb 22, 2010

### Staff: Mentor

Re: Sequences

No, that won't do. It's true that n + 1 is always > n (assuming n >0), but 1/(n + 1) < 1/n. If each expression on the left side was larger that the corresponding expression on the right side, then I would buy it.

How do you know that for n = 37, or 503, or whatever, that n + 1 + 1/(n + 1) isn't less than n + 1/n?

8. Feb 22, 2010

### Ki-nana18

Re: Sequences

Thank you for all your help. But I am completely lost. But I'll try one more question. Do I solve an+1>= an, for n?

9. Feb 22, 2010

### Staff: Mentor

Re: Sequences

Yes.

10. Feb 22, 2010

### Ki-nana18

Re: Sequences

Ok, so tons of algebra. Thank you for all your help.

11. Feb 22, 2010

### Staff: Mentor

Re: Sequences

It's hardly "tons of algebra." Unless you think three of so lines constitutes "tons."

Presumably you're in a calculus class if you're asking questions about sequences, so it's reasonble to assume that you have mastered algebra to some extent.

12. Feb 22, 2010

### Ki-nana18

Re: Sequences

Is this right:
n+1+1/(n+1)≥n+(1/n)
(n+1)^2/(n+1)≥((n^2+1))/n
n(n+1)≥ ((n^2+1))/n (n)
n^2+n≥n^2+1
n≥1

13. Feb 22, 2010

### Staff: Mentor

Re: Sequences

There's a mistake in the line above, on the left side.
How does the line above follow from the line above it?
Another way you can do this is to show that n + 1 + 1/(n + 1) - (n + 1/n) ≥ 0 for all n ≥ M, where you specify what M is.

14. Feb 22, 2010

### Ki-nana18

Re: Sequences

Ok I did an+1-an>=0 and I got (n2+3n+1)/(n(n+1))>=0. I'm not sure what to do now. I can't cancel any "n" out b/c its all addition. What do I do now?

15. Feb 22, 2010

### Staff: Mentor

Re: Sequences

n + 1 + 1/(n + 1) - n - 1/n
= 1 + 1/(n + 1) - 1/n
= ?

Leave the first 1 as-is.

16. Feb 22, 2010

### Ki-nana18

Re: Sequences

So it equals (2n+1)/(n2+n)>=-1?

17. Feb 22, 2010

### Staff: Mentor

Re: Sequences

***Leave the first 1 as-is.***

Just rewrite 1 + 1/(n + 1) - 1/n as an expression it is equal to. I don't want to see any inequality sign yet.

18. Feb 22, 2010

### Ki-nana18

Re: Sequences

(n^2+3n+1)/(n(n+1))?

19. Feb 22, 2010

### Staff: Mentor

Re: Sequences

DON'T DO ANYTHING WITH THE FIRST 1!!!

1 + 1/(n + 1) - 1/n = 1 + ?

20. Feb 22, 2010

### Ki-nana18

Re: Sequences

1+(1/(n(n+1))? sorry.