Proving that $\left(1 + \frac{x}{n}\right)^n, n \in \mathbb N$ is bounded Please can you help me with the following question
$$E(x)  = \left\{ \left(1 + \frac{x}{n}\right)^n : n \in \mathbb N \right\}$$
Let a(x) = sup E(x) (least upper bound) without finding the sup of E(x)
prove that a(x) < a(y) if 0 < x < y and a(x)a(y) <= {a[ (x+y) / 2] } ^ 2 and a(x + y) = a(x).a(y)
thank you
 A: One Approach
Note that for $n\gt|x|$,
$$
\left(1+\frac xn\right)^n\left(1-\frac xn\right)^n
=\left(1-\frac{x^2}{n^2}\right)^n
\le1\tag{1}
$$
implies that
$$
\left(1+\frac xn\right)^n\le\frac1{\left(1-\frac xn\right)^n}\tag{2}
$$
Bernoulli's Inequality, proven at the end of this answer and generalized at the end of this answer, says that $\left(1-\frac xn\right)^n$ is an increasing sequence for $n\gt|x|$. Thus, the right hand side of $(2)$ is a decreasing, positive sequence for $n\gt|x|$, and therefore, bounded.

A Second Approach
Pick $k\ge\max(x,1)$. Then, for $n\ge\max(-x,1)$,
$$
\begin{align}
\left(1+\frac xn\right)^n
&\le\left(1+\frac x{kn}\right)^{kn}\tag3\\
&=\left(1-\frac x{kn+x}\right)^{-kn}\tag4\\
&\le\left(1-\frac{nx}{kn+x}\right)^{-k}\tag5\\
&\le\left(1-\frac xk\right)^{-k}\tag6
\end{align}
$$
Explanation:
$(3)$: Bernoulli's Inequality ($k\ge1$)
$(4)$: $x=(1/x)^{-1}$
$(5)$: Bernoulli ($n\ge1$)
$(6)$: $\frac{nx}{kn+x}\le\frac xk$
A: Note that $$\binom{n}{k}\frac{x^k}{n^k}=\frac{x^k}{k!}\frac{n(n-1)\cdots(n-k+1)}{n^k}\leq\frac{x^k}{k!}$$
Thus for each $n$, $(1+x/n)^n\leq\sum_{k=0}^n\frac{x^k}{k!}\leq e^x$. Thus the set $E(X)$ is bounded above by $e^x$.
