# Squeezing Theorem

#### Ki-nana18

1. The problem statement, all variables and given/known data

Use the natural logarithm function to prove that (nn/en-1)< n! <
((n+1)n+1/en). Then use the squeezing theorem to find the limit as n approches infinity of nth root of (n!) all divided by n.

2. Relevant equations

3. The attempt at a solution

Related Calculus and Beyond Homework News on Phys.org

#### I like Serena

Homework Helper
For starters you could apply "ln" to all parts of the equation.

The proofing methods that spring to mind are:
• proof by induction

Do you know what those are?
If so, did you try to apply them?

#### Ki-nana18

No, I don't know what those are, but I'll look them up and learn them.

#### Ki-nana18

I applied ln to each part of the equation then plugged in n=10 and got 14.0259<15.1044<16.3768. This proves the equation for n=10, but how do I prove it for all real numbers?

#### I like Serena

Homework Helper
I applied ln to each part of the equation then plugged in n=10 and got 14.0259<15.1044<16.3768. This proves the equation for n=10, but how do I prove it for all real numbers?
For each a and b: ln(a*b) = ln a + ln b
Furthermore, the natural logarithm is a strictly increasing function, which means that if you apply it to the parts of an inequality, the inequality sign keeps pointing in the same direction.

Can you apply that to the inequality?

#### Ki-nana18

Yeah that makes total sense. I'm gonna write explanation because I can't write it mathematically. Can you help me with the second part?

#### I like Serena

Homework Helper
Proof by full induction consists of 2 steps:
1. You proof that the inequality holds for some specific n0
2. You proof that if the inequality holds for som n >= n0, it also holds for n+1.

In our case the inequality holds for n=2 because:

$$\frac 4 e = \frac {2^2} {e^{2-1}} < 2! < \frac {(2+1)^{2+1}} {e^2} = \frac {27} {e^2}$$

So we need to prove that if it holds for n, it also holds for n+1.

If we multiply all parts of the inequality for n with (n+1) we get:

$$\frac {n^n} {e^{n-1}} (n+1) < (n+1)! < \frac {(n+1)^{n+1}} {e^n} (n+1)$$

So if we can prove that:

$$\frac {(n+1)^{n+1}} {e^{(n+1)-1}} <^? \frac {n^n} {e^{n-1}} (n+1)$$

and

$$\frac {(n+1)^{n+1}} {e^n} (n+1) <\limits^? \frac {(n+2)^{n+2}} {e^{n+1}}$$

then we will have completed the proof.

Can you see where this is going?

Homework Helper

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving