I am trying to prove that this equality is correct.
\begin{equation}
\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n =\lim_{n \to\infty} \left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots + \frac{1}{n!}\right)
\end{equation}
I know that left and right hand side is equal to $e$, but as mathematician I would like to see is this two definition of $e$ is equal.
My idea is to try avoid using hard calculus tools, and try two show that two sequence has the same limit.
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1https://math.stackexchange.com/questions/69806/prove-the-definitions-of-e-to-be-equivalent – Minus One-Twelfth Dec 08 '19 at 11:54
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Have you tried binomial theorem? When you write $n!$ you assumed $n\in\mathbb N_0$ – wilsonw Dec 08 '19 at 12:59
1 Answers
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As a hint you might start by proving that: $$ \left(1+{1\over n}\right)^n \le \sum_{k=0}^n{1\over k!} \le \left(1 + {1\over n }\right)^{n+1} $$
After which the result follows, since LHS is monotonically increasing, RHS is decreasing, both LHS and RHS are bounded hence: $$ e = \lim_{n\to\infty}\left(1+{1\over n}\right)^n \le \lim_{n\to\infty}\sum_{k=0}^n{1\over k!} \le \lim_{n\to\infty}\left(1 + {1\over n }\right)^{n+1} = e $$
roman
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