# Proving that a function is monotonic and bounded

1. Dec 9, 2008

2. Dec 9, 2008

### PingPong

I think you may be making the monotone part a bit too complicated. Look at $$x_{n+1}/x_n$$. What can you tell about this ratio?

For the bounded part, why try to look at an upper bound for the numerator and a lower bound for the denominator?

3. Dec 9, 2008

### Focus

Your series is $$x_n=\prod_{i=1}^n \frac{i+9}{2i-1}$$. I'll give you a hint, $$\frac{i+9}{2i-1}=\frac{1}{4} \frac{19}{2i-1}$$. What happens as i increases (after i=5)?

Good luck

4. Dec 10, 2008

### transgalactic

why i need to prove that An+1<An
??

5. Dec 10, 2008

### PingPong

If you can show that A_(n+1)<A_n for n greater than some value, then it's decreasing for large enough n - which means its monotonic.

6. Dec 10, 2008

### Staff: Mentor

Is there a typo in the line above? I don't see how this could be true.

7. Dec 12, 2008

### Focus

Yeah sorry that should be$$\frac{i+9}{2i-1}=\frac{1}{2}+\frac{1}{4} \frac{19}{2i-1}$$

8. Dec 12, 2008

### Staff: Mentor

I don't see that those two expressions are equal, either. On the right side you get 4i - 17 in the numerator, and 4(2i - 1) in the denominator.

9. Dec 15, 2008

### Focus

Yeah right again I miscalculated the factor $$\frac{i+9}{2i-1}=\frac{1}{2}+\frac{1}{2} \frac{19}{2i-1}$$

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