Prove Squeezing Theorem with Natural Log: Find Limit as n->∞

[Ki-nana18]

Homework Statement

Use the natural logarithm function to prove that (nⁿ/e^n-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.

Homework Equations

The Attempt at a Solution

I'm not sure where to even start. Please help.

 

[I like Serena]

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

The proofing methods that spring to mind are:

• proof by induction
• proof by contradiction

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]

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?

Let's start with the definition of the natural logarithm.
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 going to write explanation because I can't write it mathematically. Can you help me with the second part?

 

[I like Serena]

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}} &lt; 2! &lt; \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) &lt; (n+1)! &lt; \frac {(n+1)^{n+1}} {e^n} (n+1)

So if we can prove that:

\frac {(n+1)^{n+1}} {e^{(n+1)-1}} &lt;^? \frac {n^n} {e^{n-1}} (n+1)

and

\frac {(n+1)^{n+1}} {e^n} (n+1) &lt;\limits^? \frac {(n+2)^{n+2}} {e^{n+1}}

then we will have completed the proof.

Can you see where this is going?

 

[I like Serena]

If you don't like full induction, here's an alternative using integrals and the natural logarithm:

http://en.wikipedia.org/wiki/N!#Rate_of_growth