Exponential as Limit of Sequence

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Let $e$ be defined as in Euler's number as the number satisfied by $\ln e = 1$.

Let $\left \langle {x_n} \right \rangle$ be the sequence in $\R$ defined as:

$x_n = \left({1 + \dfrac x n}\right)^n$

Then $\left \langle {x_n} \right \rangle$ converges to the limit $e^x$


$\displaystyle \lim_{n \to \infty} \left({1 + \frac 1 n}\right)^n = e$

Proof of Convergence

From Equivalence of Definitions of Exponential Function, we have:

$\displaystyle \lim_{n \to \infty} \left({1 + \dfrac x n}\right)^n = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$

the latter of which converges from Series of Power over Factorial Converges.


Proof of Convergence to $e^x$

This proof assumes the Laws of Logarithms.

We have:

\(\displaystyle \ln \left({\left({1 + \frac x n}\right)^n}\right)\) \(=\) \(\displaystyle n \ln \left({1 + xn^{-1} }\right)\) $\quad$ Logarithms of Powers $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x \frac {\ln \left({1 + x n^{-1} }\right)} {x n^{-1} }\) $\quad$ multiplying by $1 = \dfrac { xn^{-1} }{ xn^{-1} }$ $\quad$

From Limit of Sequence is Limit of Real Function, we can consider the differentiable analogue of the sequence.

From Derivative of Logarithm at One we have:

$\displaystyle \lim_{x \to 0} \frac {\ln \left({1 + x}\right)} x = 1$

But $x n^{-1} \to 0$ as $n \to \infty$ from Power of Reciprocal.


$\displaystyle x \frac {\ln \left({1 + x n^{-1}}\right)} {x n^{-1}} \to x$

as $n \to \infty$.

From Exponential Function is Continuous:

$\displaystyle \left({1 + \frac x n}\right)^n = \exp \left({n \ln \left({1 + \frac x n}\right)}\right) \to \exp x = e^x$

as $n \to \infty$.


Proof of Corollary

From Equivalence of Definitions of Euler's Number:

$e = e^1$

The result follows by setting $x = 1$ in the main result.