Prove By Induction N^(n+1) > (n+1)^n For N ≥ 3

Hey guys! Today, we're diving deep into a fascinating problem using the power of mathematical induction. We're going to prove that for all integers n greater than or equal to 3, the inequality n^(n+1) > (n+1)^n holds true. This is a classic example of how induction can be used to tackle inequalities involving exponents. So, grab your thinking caps, and let's get started!

Understanding Mathematical Induction

Before we jump into the specifics of this problem, let's quickly recap what mathematical induction is all about. At its core, mathematical induction is a powerful technique for proving statements that hold for all natural numbers greater than or equal to some initial value. It's like setting up a chain reaction – if you can show that the first domino falls and that each domino knocks over the next one, then you know that all the dominos will fall.

The process of mathematical induction involves three main steps:

1. Base Case: You need to show that the statement is true for the smallest value in your range (in our case, n = 3). This is the first domino that needs to fall.
2. Inductive Hypothesis: You assume that the statement is true for some arbitrary integer k within your range. This is like assuming that a domino at position k falls.
3. Inductive Step: You need to prove that if the statement is true for k, then it must also be true for k + 1. This is the crucial step where you show that if the domino at position k falls, it will knock over the domino at position k + 1.

Once you've successfully completed these three steps, you've proven that the statement holds true for all integers in your range. It's a pretty neat trick, right?

Proof by Induction

Now that we've refreshed our understanding of mathematical induction, let's apply it to our problem: proving that n^(n+1) > (n+1)^n for all n ≥ 3.

1. Base Case (n = 3)

First, we need to show that the inequality holds true for the base case, which is n = 3. Let's plug in n = 3 into our inequality:

3^(3+1) > (3+1)^3

3^4 > 4^3

81 > 64

As you can see, 81 is indeed greater than 64, so the inequality holds true for n = 3. This means our first domino has fallen!

2. Inductive Hypothesis

Next, we assume that the inequality holds true for some arbitrary integer k ≥ 3. This means we assume that:

k^(k+1) > (k+1)^k

This is our inductive hypothesis. We're assuming that this statement is true, and we're going to use this assumption to prove that the statement is also true for k + 1.

3. Inductive Step

Now comes the heart of the proof: the inductive step. We need to show that if our inductive hypothesis is true (i.e., k^(k+1) > (k+1)^k), then the inequality must also hold true for n = k + 1. In other words, we need to prove that:

(k+1)^((k+1)+1) > ((k+1)+1)^(k+1)

(k+1)^(k+2) > (k+2)^(k+1)

This is what we need to prove. Let's start with the left-hand side of the inequality, (k+1)^(k+2), and try to manipulate it to show that it's greater than the right-hand side, (k+2)^(k+1).

We can rewrite (k+1)^(k+2) as:

(k+1)^(k+2) = (k+1)^k * (k+1)^2

Now, remember our inductive hypothesis? We assumed that k^(k+1) > (k+1)^k. We can rearrange this inequality to get:

(k+1)^k < k^(k+1)

Let's substitute this into our expression:

(k+1)^(k+2) = (k+1)^k * (k+1)^2 < k^(k+1) * (k+1)^2

Now, we need to show that k^(k+1) * (k+1)^2 is greater than (k+2)^(k+1). This is where things get a little tricky, but we can use a clever trick. Let's divide both sides of the inequality we're trying to prove by (k+1)^(k+1). This gives us:

((k+1)^(k+2)) / ((k+1)^(k+1)) > ((k+2)^(k+1)) / ((k+1)^(k+1))

Which simplifies to:

k + 1 > ((k+2) / (k+1))^(k+1)

Now, let's focus on the right-hand side of the inequality. We can rewrite ((k+2) / (k+1)) as (1 + 1/(k+1)). So, our inequality becomes:

k + 1 > (1 + 1/(k+1))^(k+1)

Here's the crucial part: we know that the expression (1 + 1/n)^n approaches e (Euler's number, approximately 2.71828) as n approaches infinity. This is a well-known limit in calculus.

Since k ≥ 3, we know that k + 1 is also greater than or equal to 4. Therefore, (1 + 1/(k+1))^(k+1) will be less than e, which is less than 3.

So, we have:

k + 1 > (1 + 1/(k+1))^(k+1)

Since k >=3, k + 1 >= 4 > e > (1 + 1/(k+1))^(k+1)

This proves that (k+1)^(k+2) > (k+2)^(k+1).

Conclusion

We have successfully completed all three steps of mathematical induction:

1. We showed that the inequality holds true for the base case, n = 3.
2. We assumed that the inequality holds true for some arbitrary integer k ≥ 3.
3. We proved that if the inequality holds true for k, then it also holds true for k + 1.

Therefore, by the principle of mathematical induction, we have proven that the inequality n^(n+1) > (n+1)^n holds true for all integers n ≥ 3.

That was quite a journey, guys! We tackled a challenging inequality using the power of mathematical induction. Remember, induction is a powerful tool for proving statements about natural numbers, and it's all about setting up that chain reaction. Keep practicing, and you'll become masters of induction in no time!

Keywords: Mathematical Induction, Inequality, Exponents, Base Case, Inductive Hypothesis, Inductive Step