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.

Proof
See Exponential as Limit of 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:

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 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.

Also see

 * Equivalence of Exponential Definitions