an expression for the $ e^x $ using the binomial theorem

Question

Is it possible using the Binomial theorem , to prove the identity

$$ e \sim \left(1+\frac{1}{n}\right)^\frac{1}{n}\sim \sum_{k=0}^n\frac{1}{k!} $$

where $ n \to \infty $

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\pars{1 + {1 \over n}}^{n}} = \sum_{k = 0}^{n}{n \choose k} \pars{1 \over n}^{k} = \sum_{k = 0}^{n}{1 \over k!} \,{n! \over \pars{n - k}!n^{k}} \\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\hspace{3mm} \sum_{k = 0}^{n}{1 \over k!} \,{\root{2\pi}n^{n + 1/2}\,\,\expo{-n} \over \bracks{\root{2\pi} \pars{n - k}^{n - k + 1/2}\,\,\, \expo{-\pars{n - k}}\,}n^{k}} \\[5mm] = & \ \sum_{k = 0}^{n}{1 \over k!}\,{\expo{-k} \over \bracks{n^{-k}\,\,\pars{1 - k/n}^{n}\,\, \pars{1 - k/n}^{1/2 - k}\,\,}n^{k}} \\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim} & \hspace{5mm} \sum_{k = 0}^{n}{1 \over k!}\,{\expo{-k} \over \bracks{n^{-k}\ \times\ \expo{-k}\ \times\ 1} n^{k}} \\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\to} & \hspace{5mm} \bbox[#ffe,15px,border:1px dotted navy]{\ds{\color{#44f}{\sum_{k = 0}^{n}\,\,{1 \over k!}}}} \\ & \end{align}