Proving that a function is monotonic and bounded

[transgalactic]

in this link i written the question and how i tried to solve them

http://img504.imageshack.us/img504/7371/95405842kw4.gif

how to finish it??

 

[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?

 

[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

 

[transgalactic]

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

 

[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.

 

[Mark44]

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}.

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

What happens as i increases (after i=5)?

Good luck

 

[Focus]

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

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

 

[Mark44]

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

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.

 

[Focus]

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.

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