# Understanding Natural Numbers and Bernoulli's Inequality

• cooljosh2k2
In summary: 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.In summary, the homework statement is trying to prove that if n belongs to N, and:An: = (1 + 1/n)^nthen An < An+1 for all natural n.
cooljosh2k2

## 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 I am sure induction is involved in this proof. However, I am 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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If an >a(n+1), what does that imply about an / a(n+1)

fuzzywolf said:
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

Ah, then if an < a(n+1), an/an+1 < 1 ... Have you tried showing that inequality?

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? I am not sure where to go from here

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Any help guys?

cooljosh2k2 said:
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.

Dick said:
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 I am 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.

cooljosh2k2 said:
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 I am 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.

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 I am stuck. Did i go about this the wrong way, or is there something that I am missing?

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.

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 I am wrong here. I don't understand what other type of algebra manipulation I am supposed to do, and don't understand how to find r.

cooljosh2k2 said:
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 I am wrong here. I don't understand what other type of algebra manipulation I am supposed to do, and don't 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.

Dick said:
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 >= ?

cooljosh2k2 said:
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?

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?

cooljosh2k2 said:
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?

Dick said:
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. I am not the brightest guy when it comes to this, but I am 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))?

cooljosh2k2 said:
Huh? sorry you've lost me. I am not the brightest guy when it comes to this, but I am 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?

Right, that's 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.

cooljosh2k2 said:
Right, that's 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.

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?

cooljosh2k2 said:
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.

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?

cooljosh2k2 said:
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.

Thanks sooooooo much :)

## 1. What are natural numbers?

Natural numbers, also known as counting numbers, are a set of positive integers that are used for counting and ordering objects. They start from 1 and continue in an infinite sequence, with each number being one more than the previous number. Some examples of natural numbers are 1, 2, 3, 4, 5, etc.

## 2. What is the significance of natural numbers in mathematics?

Natural numbers have a significant role in mathematics as they are the foundation of many mathematical concepts and operations. They are used in basic arithmetic operations such as addition, subtraction, multiplication, and division. They also serve as the building blocks for more complex number systems, such as integers, rational numbers, and real numbers.

## 3. What is Bernoulli's inequality?

Bernoulli's inequality is a mathematical statement that shows the relationship between exponents and inequalities. It states that for any real number x greater than -1 and any positive integer n, the inequality (1+x)^n ≥ 1 + nx holds. This inequality is named after the Swiss mathematician Jacob Bernoulli who first discovered it.

## 4. How is Bernoulli's inequality used in mathematics?

Bernoulli's inequality is used in various mathematical proofs and calculations, especially in the field of calculus and analysis. It is also used in probability theory, where it is used to prove the law of large numbers and the central limit theorem. In addition, it has applications in economics, physics, and engineering.

## 5. Can Bernoulli's inequality be extended to other exponents?

Yes, Bernoulli's inequality can be extended to other exponents beyond positive integers. For example, it can be extended to rational exponents using the binomial theorem. It can also be extended to real and complex numbers using the concept of limits. However, the inequality may not hold for all values of x and n in these cases.

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