Can you prove n+1 ≥ (1+1/n)^n using induction?

[quasar987]

Basically I will have won if I can show that

$$n+1 \geq \left( 1+ \frac{1}{n} \right)^n$$

can I? Not that I haven't tried...

This is also equivalent to showing that

$$n^n \geq (n+1)^{n-1}$$

 

[quasar987]

I found it. It wasn't so easy but it turns out that the sequence on the right has 3 for an upper bound. More precisely, it has e, the neperian number, has its supremum.

 

[M98Ranger]

for all positive integers n. Yes, you can prove this using induction.

First, the base case n = 1:

1^1 \geq (1+1)^{1-1}

1 \geq 2^0

1 \geq 1

which is true.

Next, assume that the statement is true for some positive integer k, meaning

k^k \geq (k+1)^{k-1}

Now, we want to show that the statement is also true for k+1:

(k+1)^{k+1} \geq (k+2)^k

Expanding the left side, we get:

(k+1)^{k+1} = (k+1) \cdot (k+1)^k

Using our assumption, we can replace (k+1)^k with k^k \cdot (k+1), giving us:

(k+1)^{k+1} = (k+1) \cdot k^k \cdot (k+1)

Simplifying, we get:

(k+1)^{k+1} = k^{k+1} \cdot (k+1)^2

Now, we can use the fact that k \geq 1 to say that k+1 \geq 2. Therefore, (k+1)^2 \geq k+2, and we can replace it in our equation:

(k+1)^{k+1} \geq k^{k+1} \cdot (k+2)

Finally, using our assumption again, we can say that k^k \geq (k+1)^{k-1}, so:

(k+1)^{k+1} \geq (k+1)^{k-1} \cdot (k+2)

Which is true, since we are assuming k+1 \geq 2.

Therefore, by induction, the statement is true for all positive integers n. This means that n+1 \geq \left( 1+ \frac{1}{n} \right)^n is also true for all positive integers n, and you have proven your statement.