Show that if n belongs to N, and: An: = (1 + 1/n)^n then An > An+1 for all natural n

  • #1

Homework Statement



Show that if n belongs to N, and:

An: = (1 + 1/n)^n

then An < An+1 for all natural n. (Hint, look at the ratios An+1/An, and use Bernoulli's inequality)

The Attempt at a Solution



I think i have a vague idea of what to do here, like im sure induction is involved in this proof. However, im unsure how bernoullis inequality and the ratios help in the proof, and how they should be used. Can anyone help me please?
 
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Answers and Replies

  • #2
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If an >a(n+1), what does that imply about an / a(n+1)
 
  • #3


If an >a(n+1), what does that imply about an / a(n+1)

It would imply that an/an+1 is > 1, however, i screwed up the wording of the problem, its supposed to be an < an+1, and an+1/an
 
  • #4
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Ah, then if an < a(n+1), an/an+1 < 1 ..... Have you tried showing that inequality?
 
  • #5


How would i show the inequality an+1/an > 1. Would it just be: (1 + 1/(n+1))^(n+1)/(1+1/n)^n > 1?

Then (1+1/n+1)^(n+1) > (1+1/n)^n, which would then give, (1+1/n+1)*(1+1/n+1)^n > (1+1/n)^n.

Is this correct? Im not sure where to go from here
 
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  • #7
Dick
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Any help guys?

Try and use more parentheses if you aren't going to use TeX. That's not only hard to read, it's ambiguous. It looks to me like you've gotten up to wanting to prove (1+1/(n+1))*((1+1/(n+1))/(1+1/n))^n>1. It's not clear what to do with that, but then you haven't used Bernoulli's inequality like the hint suggests. Try and apply it to the second factor by doing some algebra.
 
  • #8


Try and use more parentheses if you aren't going to use TeX. That's not only hard to read, it's ambiguous. It looks to me like you've gotten up to wanting to prove (1+1/(n+1))*((1+1/(n+1))/(1+1/n))^n>1. It's not clear what to do with that, but then you haven't used Bernoulli's inequality like the hint suggests. Try and apply it to the second factor by doing some algebra.


I apologize for my crappy typing. What is TeX and how do you use it? I want to make it as clear as possible, but for now using more parenthesis, the inequality im looking for is:

[(1+1/(n+1))((1+1/(n+1)^n)] / (1+1/n)^n > 1

How do you apply bernoullis inequality, i know the basic concept of it, but am not sure how to apply it.
 
  • #9
Dick
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I apologize for my crappy typing. What is TeX and how do you use it? I want to make it as clear as possible, but for now using more parenthesis, the inequality im looking for is:

[(1+1/(n+1))((1+1/(n+1)^n)] / (1+1/n)^n > 1

How do you apply bernoullis inequality, i know the basic concept of it, but am not sure how to apply it.

Let's skip the TeX for now, I know what you are saying. That's basically the same thing I wrote except I combined the two nth power terms into one fraction. Do you see that? Now you want to do some algebra on that fraction so you can write it as (1+r)^n. What is the r part? Then it should be clear how to apply Bernoulli to the nth power.
 
  • #10


Oh i see what you did there, didnt notice it at first. The algebra is throwing me off now, i found a common denominator for both the numerator and denominator for the fraction to the nth:

[(n+2)/(n+1)] / [(n+1)/(n)]^n

which gave n^2 + 2n / n^2 + 2n +1

but now im stuck. Did i go about this the wrong way, or is there something that im missing?
 
  • #11
Dick
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Check your algebra. It looks like you turned (1+1/(n+1))/(1+1/n) into n/(n+1). That's not right. And I don't see how you 'got' (n^2+2n)/(n^2+2n+1). Got for what? I just want you to figure out what r is if 1+r=(1+1/(n+1))/(1+1/n). Then you can apply Bernoulli's inequality to (1+r)^n.
 
  • #12


What i did was:

(1 + 1/(n+1))/(1+1/n) ---->(find common denominator for top and bottom fractions)

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

(then i multiplied it out to get: --> (n^2 + 2n) / (n^2 + 2n +1)

clearly im wrong here. I dont understand what other type of algebra manipulation im supposed to do, and dont understand how to find r.
 
  • #13
Dick
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What i did was:

(1 + 1/(n+1))/(1+1/n) ---->(find common denominator for top and bottom fractions)

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

(then i multiplied it out to get: --> (n^2 + 2n) / (n^2 + 2n +1)

clearly im wrong here. I dont understand what other type of algebra manipulation im supposed to do, and dont understand how to find r.

That's fine. I just didn't understand what (n^2 + 2n) / (n^2 + 2n +1) was supposed to be from your post. Now you want to write (n^2 + 2n) / (n^2 + 2n +1) in the form 1+r. Just set 1+r=(n^2 + 2n) / (n^2 + 2n +1) and solve for r.
 
  • #14


That's fine. I just didn't understand what (n^2 + 2n) / (n^2 + 2n +1) was supposed to be from your post. Now you want to write (n^2 + 2n) / (n^2 + 2n +1) in the form 1+r. Just set 1+r=(n^2 + 2n) / (n^2 + 2n +1) and solve for r.

Ok, i get r = -1 / (n^2 + 2n + 1 ), so then what would be my inequality?

[1 -1 / (n^2 + 2n + 1 )]^ n >= ?
 
  • #15
Dick
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Ok, i get r = -1 / (n^2 + 2n + 1 ), so then what would be my inequality?

[1 -1 / (n^2 + 2n + 1 )]^ n >= ?

Good. Now what does Bernoulli's inequality say (1+r)^n is greater than? And is Bernoulli's inequality valid for this value of r? You did look up Bernoulli's inequality, right?
 
  • #16


Yes, i have bernoullis inequality right in front of me:

(1+x)^n ≥ 1+ nx for all natural numbers.

So would it be:

[1 -1 / (n^2 + 2n + 1 )]^n ≥ 1 - n / (n^2 + 2n + 1 )

and what happens to the other 1+1/(n+1) from the original equation?
 
  • #17
Dick
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Yes, i have bernoullis inequality right in front of me:

(1+x)^n ≥ 1+ nx for all natural numbers.

So would it be:

[1 -1 / (n^2 + 2n + 1 )]^n ≥ 1 - n / (n^2 + 2n + 1 )

and what happens to the other 1+1/(n+1) from the original equation?

You've got that your original expression is greater than or equal to (1 - n / (n^2 + 2n + 1 ))*(1+1/(n+1)). If you can show that is greater than one, then you done, right?
 
  • #18


You've got that your original expression is greater than or equal to (1 - n / (n^2 + 2n + 1 ))*(1+1/(n+1)). If you can show that is greater than one, then you done, right?

Huh? sorry you've lost me. Im not the brightest guy when it comes to this, but im really trying to understand it. So i have:

(original-->) (1 + 1/(n+1))/(1+1/n) >= (1 - n / (n^2 + 2n + 1 ))*(1+1/(n+1))?
 
  • #19
Dick
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Huh? sorry you've lost me. Im not the brightest guy when it comes to this, but im really trying to understand it. So i have:

(original-->) (1 + 1/(n+1))/(1+1/n) >= (1 - n / (n^2 + 2n + 1 ))*(1+1/(n+1))?

Nope. a_n+1/a_n=(1+1/(n+1))^(n+1)/(1+1/n)^n >= (1 - n / (n^2 + 2n + 1 ))*(1+1/(n+1)). Go back through the steps. We originally wanted to show a_n+1/a_n>1 to show a_n+1>a_n. Remember?
 
  • #20


Right, thats what i meant to type:

(1+1/(n+1))^(n+1)/(1+1/n)^n >= (1 - n / (n^2 + 2n + 1 )^n)*(1+1/(n+1))

So what do i do next :S. I really appreciate your help by the way.
 
  • #21
Dick
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Right, thats what i meant to type:

(1+1/(n+1))^(n+1)/(1+1/n)^n >= (1 - n / (n^2 + 2n + 1 )^n)*(1+1/(n+1))

So what do i do next :S. I really appreciate your help by the way.

You are almost done! Now you just have to show the expression on the right is greater than 1. Just multiply it out and put everything over a common denominator. See if it looks greater than 1.
 
  • #22


Ok, if i multiply the entire right hand side, i get (n^3 + 3n^2 + 3n + 2) / (n^3 + 3n^2 + 3n + 1). Which is greater than 1, so does that prove that (1+1/(n+1))^(n+1)/(1+1/n)^n > 1, and therefore an+1 > an?
 
  • #23
Dick
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Ok, if i multiply the entire right hand side, i get (n^3 + 3n^2 + 3n + 2) / (n^3 + 3n^2 + 3n + 1). Which is greater than 1, so does that prove that (1+1/(n+1))^(n+1)/(1+1/n)^n > 1, and therefore an+1 > an?

By now you should be able to explain to me why it does.
 
  • #24


Since the right hand side is greater than 1, and the left hand side is greater than or equal to the right hand side, then the left hand side must be greater than 1. In which case, an+1 > an. Is this good?
 
  • #25
Dick
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Since the right hand side is greater than 1, and the left hand side is greater than or equal to the right hand side, then the left hand side must be greater than 1. In which case, an+1 > an. Is this good?

Sure.
 

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