Show if the sequence converges

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In summary, the sequence given by the equation n/(2^n) is decreasing with a lower bound. The steps to prove this include finding the sequence, using the inequality an+1 < an, simplifying and comparing the right and left sides, and recognizing that the sequence is underbounded. The process may be confusing at first, but with practice it can be understood and applied to other problems.
  • #1
vipertongn
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Homework Statement


showing it is increasing with an upper bound or decreasing
with a lower bound

n/(2^n)

Homework Equations



if an >an+1 its underbound
an< an+1 then its upperbound

The Attempt at a Solution


I tried first by finding out the sequence:
a1=1/2
a2=1/2
a3=3/8
a4=1/4
a5=5/32

I'm assuming its decreasing, but I'm not sure if this is monotonic at all. Considering how a1=a2 and then a2>a3 and a3>a4 then a4>a5. I think it's underbound since an>an+1 but the first part threw me off since a1=a2. Someone clarify for me?
 
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  • #2
It is really trivial to show that sequence is decreasing. Have you tried?
 
  • #3
Yea I can tell that the sequence is decreasing just by looking at that. I was taught to try (an+1)-an see if that's greater than 0 or not, but its kinda confusing with this equation. then there is also taking the derivative but it also doesn't help much because that ends up being just 1/2^n-1/2^n(log2)
 
  • #4
vipertongn said:
Yea I can tell that the sequence is decreasing just by looking at that. I was taught to try (an+1)-an see if that's greater than 0 or not, but its kinda confusing with this equation.

That's good advice. What happens if you work on the inequality an+1 < an with reversible steps? Try putting the 2's on one side by themselves.
 
  • #5
LCKurtz said:
That's good advice. What happens if you work on the inequality an+1 < an with reversible steps? Try putting the 2's on one side by themselves.

Ummm... do you think you could give me an example as to how to go about this. I have other problems and this one seems the most simple to actually work with. >.<;; my professor doesn't give good examples and that's usually how I work through my math. Let's see if I can get this right

so n+1/(2n+1< n/2n
That turns into n+1/n < 2n+1/2n If i subtract that...
so...
n+1/n-2n+1/2n<0 so that shows that its decreasing and underbounded (meaning its bounded under something right?

but wait if i do it the other way... isn't 0>2n+1/2n-n+1/n
 
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  • #6
vipertongn said:
Ummm... do you think you could give me an example as to how to go about this. I have other problems and this one seems the most simple to actually work with. >.<;; my professor doesn't give good examples and that's usually how I work through my math. Let's see if I can get this right

so n+1/(2n+1< n/2n
That turns into n+1/n < 2n+1/2n If i subtract that...

You need parentheses around the n+1 on the left don't you? Don't subtract anything. Just look at what you have. Simplify the right side. Compare it to the left.
 
  • #7
LCKurtz said:
You need parentheses around the n+1 on the left don't you? Don't subtract anything. Just look at what you have. Simplify the right side. Compare it to the left.

how do i simplify 2n+1/1n? does it become 1 or something?
 
  • #8
vipertongn said:
how do i simplify 2n+1/1n? does it become 1 or something?

[tex]\frac{2^{n+1}}{2^n}\neq\frac{2^{n+1}}{1^n}[/tex]
 

FAQ: Show if the sequence converges

What is the definition of a convergent sequence?

A sequence is said to be convergent if its terms approach a specific value as the index of the sequence increases. This value is called the limit of the sequence.

How can I determine if a sequence converges?

To determine convergence, we can use several methods such as the limit comparison test, root test, or ratio test. We can also check if the sequence is monotonic and bounded.

What is the difference between a convergent and a divergent sequence?

A convergent sequence has a finite limit, while a divergent sequence does not. This means that the terms of a convergent sequence get closer and closer to a specific value, while the terms of a divergent sequence do not.

Can a sequence converge to more than one limit?

No, a sequence can only have one limit. If a sequence converges to more than one limit, it is considered divergent.

What is the importance of determining convergence in a sequence?

Determining convergence is important in many mathematical and scientific applications. It allows us to understand the behavior and properties of a sequence, and it helps us make predictions and draw conclusions about the values of the sequence. Convergence is also a fundamental concept in calculus and real analysis.

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