How to prove that a certain set has no largest element and is bounded above ? Isn't that a contradiction? How can I prove that the set $\left\{ \left(1 + \frac{x}{n}\right)^n\right\}$ has no largest element and is bounded above ? Isn't that a contradiction?
I don't understand how a set can be bounded above without having a largest element. I learned that by completeness every set that is bounded above has a largest element, isn't that correct? 
Finally can you please help me solving this problem or at least give me some hint? 
I would be very grateful if you helped me. Thank you :)
 A: Check out the set $A$ of numbers $a_n=1-\frac 1n$ for $n\in \mathbb N$ - each number is less than $1$, so it is bounded above.
Suppose $a_m$ were a greatest element. Now consider $a_{m+1}\gt a_m$. So there is no greatest element.
Completeness ensures that every set which is bounded above has a least upper bound - for $A$ this is $1$, but we don't have $1\in A$.
A set is said to be closed if it contains all its limit points. The least upper bound is a limit point, so we can say that every closed set which is bounded above has a greatest element.

To deal with the substance of the question using the binomial expansion, here are some pointers.
First the typical term in the expansion of $(1+\frac xn)^n$ is $\binom nr \left(\frac xn\right)^r1^{n-r}$ and we can write this as
$$\frac {n(n-1)(n-2)(n-r+1)}{n^rr!}x^r$$
or (isolating the term with $n$ in, which we want to control)
$$\left(1-\frac 1n\right)\left(1-\frac 2n\right) \ldots \left(1-\frac{r-1}n\right)\frac {x^r}{r!}\lt\frac {x^r}{r!}$$
because all the factors which have been dropped are positive and less than $1$.
So you have a potential comparison, but the series still needs to be bounded above. This can be done, for example, by showing that the first $N$ terms of the comparison series are constant (don't depend on $n$), so their sum is fixed, and that for a suitably chosen $N$ (chosen in relation to $x$, which is a fixed number) the later terms are less than the equivalent terms of a suitably chosen geometric progression, so the sum of the remaining terms can be bounded.
I've left some gaps to fill in, and you will have to take care in justifying each step.
A: Since some of your questions have been disposed of, let's focus on the actual task at hand (proving existence of an upper bound). 
Let's take the case $x=1$ first. I presume you have had some calculus, and therefore would know how to analyze a function like $f(t) = (1 + \frac1{t})^t$. For example, you should be able to determine where the function is increasing (first derivative test), whether the function has a horizontal asymptote (compute $\lim_{t \to \infty} f(t)$ using L'hôpital's rule), etc. 
Using this, you should be able to determine that the sequence $(1 + \frac1{n})^n$ increases with integers $n \geq 1$, and has an upper bound. You should even be able to give the least upper bound, based on your analysis. 
Now, for each fixed $x$, do the same with the sequence $(1 + \frac{x}{n})^n$, treating the values $\frac{n}{x}$ as special values of $t$ as $t$ tends to $\infty$. It might help to write first $(1 + \frac{x}{n})^n = ((1 + \frac{x}{n})^{\frac{n}{x}})^x$. This should be enough of a hint for your problem. 
A: I think you need to have $x > 0$ here.
You can see that
\begin{align}
s_{n} &= \left(1 + \frac{x}{n}\right)^{n}\notag\\
&= 1 + n\cdot\frac{x}{n} + \frac{n(n - 1)}{2!}\left(\frac{x}{n}\right)^{2} + \cdots\notag\\
&= 1 + x + \dfrac{1 - \dfrac{1}{n}}{2!}x^{2} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}x^{3} + \cdots\notag
\end{align}
In the above binomial expansion the number of terms is $(n + 1)$ and you can see that if $n$ increases then the value of each term increases as well as the number of terms also increases. Hence $s_{n + 1} > s_{n}$ for all positive integers $n$ and $x > 0$. This shows that the sequence has no greatest element.
Next problem is to show that the sequence $s_{n}$ is bounded above. This is also easy if you notice that the above binomial expansion leads to $$s_{n} \leq 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots + \frac{x^{n}}{n!}$$ The series $\sum x^{n}/n!$ is convergent (for example by ratio test) and since $x$ is positive any finite number of terms will always sum up to a number less than or equal to the sum of the infinite series $\sum_{n = 0}^{\infty}x^{n}/n!$. Hence $s_{n} \leq \sum_{n = 0}^{\infty}x^{n}/n!$. This proves that the sequence is bounded above.

Next let's discuss about bounded sets and greatest member. Let $A$ be a non-empty subset of $\mathbb{R}$. Suppose $A$ is bounded above so that $x \leq K$ for all $x \in A$ and some specific $K\in\mathbb{R}$. The completeness property of real numbers says that in this case there exists a real number $M$ with the following two properties:

*

*No member of $A$ exceeds $M$ (in symbols we write $x \leq M$ for all $x \in A$).

*Any number less than $M$ is always exceeded by at least one member of $A$ (in symbols we have: for any $\epsilon > 0$ there is an $x \in A$ such that $x > M - \epsilon$).

This specific number $M$ is called the least upper bound or the supremum of $A$ and written as $M = \sup A$. The number $M$ may or may belong to $A$. If $M \in A$ then $M$ is the greatest element of $A$ (proof is almost obvious) and if $M \notin A$ then $A$ has no greatest element (again the proof is obvious). You can see both the cases with the following examples $$A = \{(n - 1)/n \mid n \in \mathbb{N}\}, B = A \cup \{1\}$$ Both the sets $A, B$ are bounded above and $1 = \sup A = \sup B$ but $1 \notin A$ and $1 \in B$.
