Equivalence of Definitions of Euler's Number
Theorem
The two definitions of Euler's number $e$ as given by:
- $\displaystyle e = \lim_{n \to \infty} \left({1 + \frac 1 n}\right)^n$
- $\displaystyle e = \sum_{n=0}^\infty \frac 1 {n!}$
are equivalent to the definition of $e$ as the number satisfied by $\ln e = 1$.
Thus all definitions for $e$ are equivalent.
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Proof
See Exponential as the Limit of a Sequence for how $\displaystyle \lim_{n \to \infty} \left({1 + \frac 1 n}\right)^n = e$ follows from the definition of $e$ as the number satisfied by $\ln e = 1$.
See Euler's Number: Limit of Sequence implies Limit of Series for how $\displaystyle e = \sum_{n=0}^\infty \frac 1 {n!}$ follows from $\displaystyle \lim_{n \to \infty} \left({1 + \frac 1 n}\right)^n = e$.
Now suppose $e$ is defined as $\displaystyle e = \sum_{n=0}^\infty \frac 1 {n!}$.
Let us consider the series $\displaystyle f \left({x}\right) = \sum_{n=0}^\infty \frac {x^n} {n!}$.
From Series of Power over Factorial Converges, this is convergent for all $x$.
We differentiate $f \left({x}\right)$ WRT $x$ term by term (justified by Power Series Differentiable on Interval of Convergence), and get:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle D_x \left({f \left({x}\right)}\right)\) | \(=\) | \(\displaystyle D_x \left({1}\right) + D_x \left({\frac {x} {1!} }\right) + D_x \left({\frac {x^2} {2!} }\right) + D_x \left({\frac {x^3} {3!} }\right) + \cdots + D_x \left({\frac {x^n} {n!} }\right) + D_x \left({\frac {x^{n+1} } {\left({n+1}\right)!} }\right) + \cdots\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle 0 + 1 + \frac {2 x} {2!} + \frac {3 x^2} {3!} + \cdots + \frac {n x^{n-1} } {n!} + \frac {\left({n+1}\right) x^n} {\left({n+1}\right)!} + \cdots\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle 1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n-1} } {\left({n-1}\right)!} + \frac {x^n} {n!} \cdots\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle f \left({x}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
Thus we have $D_x \left({f \left({x}\right)}\right) = f \left({x}\right)$ and thus from Differential of Exponential Function it follows that $f \left({x}\right) = e^x$.
From Derivative of an Inverse Function we get that $D_x \left({f^{-1} \left({x}\right)}\right) = \dfrac 1 {f^{-1} \left({x}\right)}$.
Hence from Derivative of Natural Logarithm Function it follows that $f^{-1} \left({x}\right) = \ln x$.
It follows that $e$ can be defined as that number such that $\ln e = 1$.
Hence all the definitions of $e$ as given here are equivalent.
$\blacksquare$
