proof that {(1+x/n)^n} is bounded

Hello,

how can I prove that the set E(x) = { (1+x/n)^n} is bounded ? (n is a natural number and x is a positive real number)

I had already proven that the set has no largest element using a consequence of the AGM inequality.

I also know from calculus that the limit of (1+x/n)^n is actually e^x . but this is of no use to me, since e^x is not yet defined in my analysis course.

any help would be much appreciated.

Re: proof that {(1+x/n)^n} is bounded

$\displaystyle \text{You have to show that the sequence is bounded below and bounded above.}$

$\displaystyle \text{(i) \textbf{Bounded below}: Since } x \geq 0 \text{, then we have that}$

$\displaystyle 0 < \left(\frac{1+x}{n}\right)^{n} \text{ for all } n \in \mathbb{N}.$

$\displaystyle \text{(ii) \textbf{Bounded above}: Let } m \in \mathbb{N} \text{ such that }(m-1) \leq (1+x) \leq m \text{. Hence, we have that}$

$\displaystyle \frac{1+x}{m+p} < \frac{1+x}{m} \leq 1 \leq \frac{1+x}{m-1}$

$\displaystyle \implies \left(\frac{1+x}{m+p}\right)^{m+p} < \left(\frac{1+x}{m+p}\right)^{m} < \left(\frac{1+x}{m}\right)^{m} \leq 1 \leq \frac{1+x}{m-1},$

$\displaystyle \text{for all } p \in \mathbb{N} \text{. As we can see, the sequence is decreasing for all } n \geq m.$

$\displaystyle \text{Let } K > 0 \text{ be defined as}$

$\displaystyle K \equiv \max\left\{ \left(\frac{1+x}{n}\right)^{n} : 1 \leq n \leq (m-1) \right\}.$

$\displaystyle \text{Hence, we have that}$

$\displaystyle \left(\frac{1+x}{n}\right)^{n} < K \text{, for all } n \in \mathbb{N}.$

$\displaystyle \text{From (i) and (ii) we have that the sequence is bounded.} \hfill \text{Q.E.D.}$

Re: proof that {(1+x/n)^n} is bounded

hi dgomes

i think you have the wrong expression.i think his is : (1+(x/n))^n just written without the brackets.

Re: proof that {(1+x/n)^n} is bounded

Ok, you are right! At least this other example is done. I will think later on the correct one.

Did you find any mistakes on the example that I did?

Re: proof that {(1+x/n)^n} is bounded

i'm really tired,so even if there are mistakes i won't be able to catch them.but as far as i can see right now it's all pretty much correct.just to be sure you should ask someone else.