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 [itex]

0<(1+1/n)^n<3 [/itex]

Homework Equations

The Attempt at a Solution

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

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

so...

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

and...

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

But I really don't understand where to go from here do try exhaustion of cases for [itex] n, k [/itex] but that does not seem appropriate since [itex]n,k[/itex] have no boundaries other than[itex] \geq0[/itex] 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 [itex]\binom {n} {k} \leq \frac{n^k}{k!}[/itex] thus the series you have is not greater than [itex]\sum \frac{1}{k!} = e[/itex]. 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 [itex]\binom {n} {k} \leq \frac{n^k}{k!}[/itex] thus the series you have is not greater than [itex]\sum \frac{1}{k!} = e[/itex]. 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)+...