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 $<br /> 0<(1+1/n)^n<3$

Homework Equations

The Attempt at a Solution

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

$= n!/(n-k)!k! (1)^{n-k} (1/n)^{k}$ 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 $0<x\leq1$ 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)+...