Use Binomial Theorem and appropriate inequalities to prove

[charmedbeauty]

Use Binomial Theorem and appropriate inequalities to prove!

Homework Statement

Use Binomial Theorem and appropriate inequalities to prove <br /> <br /> 0&lt;(1+1/n)^n&lt;3

Homework Equations

The Attempt at a Solution

So I started by..<br /> \sum ^{n}_{k=0} (n!/(n-k)! k!) a^{n-k}b^{k}

= n!/(n-k)!k! (1)^{n-k} (1/n)^{k} for n \in Z^{+}

so...

1^{n-k} =1(since 1^{l} =1 for any l \in R)

and...

((1/n)^{k} \leq1 (Since 1/n \leq1 for any n \in Z^{+}, so from reasoning above (1/n)^{k} \leq1 and any x \in R such that 0&lt;x\leq1 then any power k\geq0 of x is going to be \leq 1)

But I really don't understand where to go from here do try exhaustion of cases for n, k but that does not seem appropriate since n,k have no boundaries other than\geq0 respectively.

Please help!

 

[hamsterman]

It is easy to prove that (1+1/n)ⁿ approaches e as n goes to infinity using L'Hôpital's rule.
If you have to do it with binomial expansion, it's more tricky.
What are "appropriate inequalities" ? I assume you've covered some in class/book ?

Here's a proof. It should be obvious that \binom {n} {k} \leq \frac{n^k}{k!} thus the series you have is not greater than \sum \frac{1}{k!} = e. This comes from Taylor series of e^x when x = 1. I don't know how to prove it without them, though.

 

[charmedbeauty]

It is easy to prove that (1+1/n)ⁿ approaches e as n goes to infinity using L'Hôpital's rule.
If you have to do it with binomial expansion, it's more tricky.
What are "appropriate inequalities" ? I assume you've covered some in class/book ?

Here's a proof. It should be obvious that \binom {n} {k} \leq \frac{n^k}{k!} thus the series you have is not greater than \sum \frac{1}{k!} = e. This comes from Taylor series of e^x when x = 1. I don't know how to prove it without them, though.

I don't think I have covered taylor series yet, maybe I better look into that. Thanks!

 

[Dick]

I don't think I have covered taylor series yet, maybe I better look into that. Thanks!

I think the comparison they want you to make is that 1+1+1/2!+1/3!+1/4!+... <= 1+1+1/2+1/(2*2)+1/(2*2*2)+...