How to find a upper bound of the set $A= \{ (1+\frac{1}{n})^n : n \in \mathbb{N}^{*} \} $ i have the set $A= \{(1+\frac{1}{n})^{n} : n \in \mathbb{N}^* \}$ and the exercise ask me to find at least $2$ upper bound least equal to $\frac{14}{5}$.
My first question is, how i know $\frac{14}{5}$ is a upper bound of $A$?
My attempt is show doesn't exist $m \in \mathbb{N}^*$ such that $(1+\frac{1}{m})^{m} = \frac{14}{5}$, but i'm stuck,  for the other upper bound numbers, i could reason the same way?
Thanks for suggestions

If you are not allowed to use the definition of $e$, then by observing that the sequence terms are bound by (refer to this)
$$
\begin{align}\left(1+\frac 1n\right)^n&=\sum_{k=0}^{n}\binom{n}{k}\left(\frac 1n\right)^k\\&=\sum_{k=0}^{n}\frac{n(n-1)\cdots(n-k+1)}{k!}\left(\frac 1n\right)^k\\&=\sum_{k=0}^{n}\frac{1}{k!}\left(1-\frac 1n\right)\left(1-\frac 2n\right)\cdots\left(1-\frac{k-1}{n}\right)\\&\lt \sum_{k=0}^{n}\frac{1}{k!}\\
&=1+\frac{1}{1}+\frac{1}{2}+\frac{1}{3\cdot 2}+\frac{1}{4\cdot 3\cdot 2}+\cdots+\frac{1}{n\cdot(n-1)\cdots 2}\\
&\lt 1+\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{n-1}}\\&=3-\frac{1}{2^{n-1}}\\&\lt 3\end{align}
$$
Now to get a stricter bound, since
$$
\frac{1}{2^1} + \frac{1}{3\cdot 2}+\frac{1}{4\cdot 3\cdot 2}+\cdots+\frac{1}{n\cdot(n-1)\cdots 2} < \frac
{1}{2} \left( 1+\frac{1}{2^0}+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{n-1}} \right)
$$
we have
$$
\left(1+\frac 1n\right)^n < 1 + \frac{1}{2^0} + \frac
{1}{2} \left(\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{n-1}} \right) < 2 + \frac{1}{2} \frac{1}{1-\frac{1}{3}} = \frac{11}{4} < \frac{14}{5}
$$
as required.
