How do I show a sequence is converging by deciding on monotonicity anf boundness?

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1. Homework Statement [/b

I want to see if a sequence converges by deciding on monotonicity and boundness.
The sequence is:

an=(n+1)/(2n+1)

How to I go about determining if it converges or not based on those two factors? I am lost on how to go about it.
THanks for any help

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The Attempt at a Solution

 

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  • #2
jbunniii
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Well, is the sequence monotone? Is it bounded? What have you tried so far?
 
  • #3
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I have tried to check on monotonicity:
(a(n+1)/an)=(2n^2+5n+2)/(2n^2 +5n +3) then multiplying by 1/n^2, I get 1 for the limit so the sequence is evnetually increasing. However, I'm not completely sure that's right and I don't know how to go about boundness
 
  • #4
jbunniii
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OK, your expression
[tex]\frac{a_{n+1}}{a_n} = \frac{2n^2 + 5n + 2}{2n^2 + 5n + 3}[/tex]
is correct. Which is larger, the numerator or the denominator?
 
  • #5
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oops, the denom is bigger so the sequence is eventually decreasing. Thanks!
How would you go about showing boundness?
 
  • #6
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I have the monotonicity for a couple of problems but don't know how to go about boundness. The one above, n/2^n, 2^n/n! etc....I just need to understand how to determine boundness :/
 
  • #7
jbunniii
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oops, the denom is bigger so the sequence is eventually decreasing.
In fact, the denominator is bigger for all [itex]n[/itex] so the sequence is in fact always decreasing, which is even better than eventually decreasing.
How would you go about showing boundness?
Quite similarly.
[tex]a_n = \frac{n+1}{2n+1}[/tex]
Which is bigger, numerator or denominator?
 
  • #8
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What does knowing which is bigger in the original problem show about boundness? So the denom being bigger in this case means what?
Thanks for all the help btw!
 
  • #9
jbunniii
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What does knowing which is bigger in the original problem show about boundness? So the denom being bigger in this case means what?
Thanks for all the help btw!
If the denominator is bigger, then how does the fraction compare to 1?
 
  • #10
HallsofIvy
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I cannot imagine why you think that a limit of 1 would mean the sequence is increasing! Both (n+1)/n, which is a decreasing sequence, and n/(n+1), which is an increasing sequence have limit 1 as n goes to infinity.

The first thing I would do is check a few values- a1= (1+ 1)/(2(1)+ 1)= 2/3, a2= (2+1)/(2(2)+ 1)= 3/5, a3= (3+1)/(2(3)+ 1)= 4/7, it certainly looks like the sequence is decreasing.

We would, then, like to say that an+1< an. Can you prove that ((n+1)+1)/(2(n+1)+1)= (n+2)/(2n+3)< (n+1)/(2n+1)? Since both 2n+ 3 and 2n+1 are positive, we can multiply both sides by (2n+3)(2n+1) to get (2n+1)(n+2)= 2n^2+ 5n+ 2 on one side and (n+1)(2n+3)= 2n^2+ 5n+ 3. It's easy to see which of those is larger!

And if, in fact, the sequence is decreasing, we only need to find a lower bound. And that should be easy! (For all n, both n+1 and 2n+ 1 are positive.)
 
  • #11
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How do you determine boundeness by comparing the fraction to 1. upper bound, lower bound, etc?
 
  • #12
jbunniii
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How do you determine boundeness by comparing the fraction to 1. upper bound, lower bound, etc?
Well, if the fraction is always smaller than 1, then isn't 1 an upper bound?

Can you find a lower bound? This is even easier.
 
  • #13
HallsofIvy
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As I pointed out, if the function is decreasing, you only need to worry about a lower bound (an upper bound is the first member of the sequence). Similarly, if a sequence is increasing, its first member is a lower bound so you would only need to show it has an upper bound.
 
  • #14
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Is that all you do to determine lower boundness? Check is numerator and denom are positive for all n>0 ?
So for a sequence such as n/2^n which is decreasing, you determing lower boundness becasue both n and 2^n are positive for all n>0?
 
  • #15
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Yeah but for this case, I have to have a lower bound
 
  • #16
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As I pointed out, if the function is decreasing, you only need to worry about a lower bound (an upper bound is the first member of the sequence). Similarly, if a sequence is increasing, its first member is a lower bound so you would only need to show it has an upper bound.
How do you find an upper bound? Sorry I' m a little new to this and confused
 
  • #17
jbunniii
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Yeah but for this case, I have to have a lower bound
[itex]n[/itex] is always positive, right? What does this imply about the fraction?
 

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