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**Use Binomial Theorem and appropriate inequalities to prove!!!**

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

Use Binomial Theorem and appropriate inequalities to prove [itex]

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

**2. Relevant equations**

**3. 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 dont 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!!

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