Calc analysis - monotone sequences

In summary, the sequence{b sub(n)} appears to be monotone decreasing with zero as the lower bound. However, to prove this, it would be necessary to show that bn+1< bn for all n.
  • #1
Nerpilis
19
0
I think I'm having some trouble on this. First I'll state what the question is then I'll show what i have and my reasoning. Determine if the sequence{b sub(n)} is convergent by deciding on monotonicity and boundness. given:
b sub(n)=n^2/2^n

First I plugged in numbers for n starting with 1. after n=3 the sequence started taking shape and it appears to be approaching zero as n gets large. my feeling at this point is that its monotone decreasing with zero as the lower bound. I guess to prove this i would have to show that b sub(n+1)< b sub(n) thus impling that the lim b sub(n) = 0 or at least exists as n goes to infinity. my first step was to show that b sub(n+1) exists by way of induction. using n=1 in the given i find it true. here is where i start to question myself. b sub(n+1) = (n+2)^2/2^(n+1). I'm not sure how to simplify this but as i plug in numbers I can easily see that 0< b sub(n+1)< b sub(n) therefore leading me to conclude that there is a lim b sub(n) = 0 as n goes to infinity. Am I flawed in this thought process, or are there holes. In my induction part are there ways of simpliying the equation? help and comments are greatly appreciated as i am rusty and many years removed from my last calc class.
 
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  • #2
I think you need to show 0 < bn+1 < bn for general n, using induction.

(n+1)^2/2^(n+1) < n^2/2^n
(n+1)^2/n^2 < 2^(n+1)/2^n
n^2 + 2n + 1 < 2n^2
etc...
 
  • #3
I'm puzzled by this: " my first step was to show that b sub(n+1) exists by way of induction."

Is there any question that n2/2n exists for all n?

Since it is also obvious that n2/2n is positive for all n (so that 0 is a lower bound), to use "monotone convergence" to show that it converges, you need only show that this is a decreasing sequence.

Generally speaking the best way to show that a sequence is decreasing (or increasing) is to compare successive terms. If bn+1< bn of all n then we must have bn+1- bn< 0 or, alternatively, bn+1/bn< 1.

In this case, because bn is defined as a fraction, I would try the latter of those: bn+1/bn= ((n+1)2/2n+1)(2n/n2)= (1/2)((n+1)2/n2). Can you show that (n+1)2/n2< 2? Well, that certainly isn't true for n= 1! What n is it true for?
 

1. What is a monotone sequence?

A monotone sequence is a sequence of numbers that either increases or decreases consistently. This means that each term in the sequence is either greater than or equal to the previous term (monotone increasing) or less than or equal to the previous term (monotone decreasing).

2. How can I tell if a sequence is monotone?

To determine if a sequence is monotone, you can compare each term to the one before it. If the terms are consistently increasing or decreasing, then the sequence is monotone. You can also graph the sequence and see if it forms a straight line or curves in a consistent direction.

3. What is the importance of monotone sequences in calculus?

Monotone sequences are important in calculus because they help us understand the behavior and limits of functions. By studying the monotonicity of a sequence, we can determine if the function is increasing or decreasing, and use this information to find the maximum and minimum values of the function.

4. Can a sequence be both monotone increasing and decreasing?

No, a sequence cannot be both monotone increasing and decreasing at the same time. It can only be one or the other. However, a sequence can alternate between being monotone increasing and decreasing, in which case it is called a oscillating sequence.

5. How is the convergence of a monotone sequence determined?

A monotone sequence is said to converge if it approaches a finite limit as the number of terms increases. The convergence of a monotone sequence can be determined by using the Monotone Convergence Theorem, which states that a bounded monotone sequence must converge to its supremum (if it is monotone increasing) or infimum (if it is monotone decreasing).

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