How to prove that $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$?

Question

How to prove that $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$ using the fact that $e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n$ ?

So, $$e^x=\lim_{n\to\infty }\sum_{k=0}^n \binom{n}{k}\left(\frac{x}{n}\right)^k=\lim_{n\to\infty }\sum_{k=0}^n\frac{n!}{(n-k)!n^k}\frac{x^k}{k!}.$$

I think that I have to prove that $\frac{n!}{(n-k)!n^k}=1$ but I didn't have success.

Answer 1

1. Solution using $\,e^{\,x} = \dfrac{d}{dx}\,e^{\,x}\,$ property of exponent

Assume that the exponent function can be represented as a series with unknown coefficients: $$ e^{\,x} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots = \sum_{n=0}^{\infty} a_n x^n $$

Recall the fundamental property of exponent $\, \dfrac{d}{dx} \big(e^{\,x} \big) = e^{\,x}$. Applying this property to the series for of exponent, we get \begin{align} \dfrac{d}{dx} \,e^{\,x} = e^{\,x} & \implies \dfrac{d}{dx} \left(\sum_{n=0}^{\infty} a_n x^n \right) = \sum_{n=0}^{\infty} a_n x^n \\ & \implies \dfrac{d}{dx} \big( a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots\big) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots \\ & \implies 0 + a_1 + 2 \, a_2\, x + 3 \, a_3\, x^2 + \ldots = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots \\ & \implies \sum_{n=1}^{\infty}n \,a_n\,x^{n-1} = \sum_{n=0}^{\infty} a_n x^n \iff \sum_{n=0}^{\infty}\left(n+1\right) a_{n+1}\,x^{n} = \sum_{n=0}^{\infty} a_n x^n \\ & \implies \left(n+1\right)a_{n+1} = a_n \\ & \implies a_{n+1} = \frac{a_n}{n+1} \end{align} The last equation can be rewritten as $\,a_{n} = \dfrac{1}{n}a_{n-1}, \,$ so that $$ a_{n} = \frac{1}{n}\,a_{n-1} = \frac{1}{n}\,\frac{1}{n-1} \,a_{n-2}= \frac{1}{n}\,\frac{1}{n-1}\,\frac{1}{n-2}\,a_{n-3} = \ldots = \frac{1}{ n!}\,a_0\tag{1.1} $$ Observer that $\,e^0 = 1,\,$ so we can write $$ e^{\,x}\Big\rvert_{x=0} = \big( a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots\big)\Big\rvert_{x=0} = a_0 = 1 $$ This fact combined with equation $(1.1)$ gives us the explicit expression for coefficient $\,a_n = \dfrac{1}{n!}.\,$ Therefore we finally write $$ \bbox[4pt, border:2.5pt solid #FF0000]{\ \ e^{\,x} = \sum_{n=0}^{\infty} \frac{x^n}{n!}\ \,} $$ Q.E.D.

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EDIT: As requested in comments, here I provide solution using $\,\displaystyle e^{\,x}=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n\,$ expression.

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2. Solution using $\,\displaystyle e^{\,x}=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n\,$ property of exponent

I do not believe that it is possible (at least within reasonable timespan) express the exponentsas $\,e^{\,x}=\sum_{k=0}^\infty \frac{x^k}{k!}\,$ using only algebraic operations and the expression formula $\, e^{\,x}=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n\,$ as starting point. However, it is possible to show the equivalence of these two definitions of $\,e^{\,x}\,$ as $\,n\to \infty.\,$

Indeed, for any $x\ge 0$ let us define $$ S_n = \sum_{k=0}^n \frac{x^k}{k!}, \qquad L_n = \left(1+\frac{x}{n}\right)^n. $$ Then, by Newton's binomial $$ \begin{aligned} L_n & = \sum_{k=0}^n {n\choose k} \,\frac{x^k}{n^k} = 1 + x + \sum_{k=2}^{n} \frac{n\cdot \left(n-1\right)\cdot \left(n-2\right)\cdot \ldots\cdot \left(n-(k-1)\right)}{k! \,n^k}= \\ & = 1 + x + \frac{x^2}{2!}\,\left(1 - \frac{1}{n} \right) + \frac{x^3}{3!}\,\left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) + \ldots + \frac{x^n}{n!}\,\left(1 - \frac{1}{n} \right) \cdots \left(1 - \frac{n-1}{n} \right) \leq S_n \end{aligned} $$ Therefore $$ \limsup_{n\to\infty}L_n \leq \limsup_{n\to\infty}S_n = e^{\,x}.\tag{2.1} $$

On the other hand, for any positive integer $\, m\,$ such that $\,2\le m \le n\,$ we have $$ 1 + x + \frac{x^2}{2!}\,\left(1 - \frac{1}{n} \right) + \ldots + \frac{x^m}{m!}\,\left(1 - \frac{1}{n} \right)\left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right) \le L_n $$ If we fix $\,m\,$ and let $\,n\to\infty,\,$ then we get $$ S_m = 1 + x + \frac{x^2}{2!} + \ldots + \frac{x^m}{m!} \leq \liminf_{n\to\infty}L_n\tag{2.2} $$ Letting $\,m\to\infty\,$ in inequality $(2.2)$ and combining it with inequality $(2.1)$, we get $$ e^{\,x} = \limsup_{n\to\infty}L_n\leq \lim_{n\to\infty} S_n \leq \liminf_{n\to\infty}L_n = e^{\,x} $$ and thus $$ \bbox[5pt, border:2.5pt solid #FF0000]{\lim_{n\to \infty}S_n = \sum_{n=0}^\infty \frac{x^n}{n!}=e^{\,x}} $$

Answer 2

To complete your method, that is, without assuming the derivative of $e^x$, you just have to write the coefficient of $\frac{x^k}{k!}$ as $$\frac{n(n-1)(n-2)...(n-k+1)(n-k)!}{(n-k)!n^k}$$ $$=(1)(1-\frac 1n)(1-\frac 2n)...(1-\frac{k-1}{n})\rightarrow1$$ as $n\rightarrow\infty$

Answer 3

My first thought would be to list defining properties of $f(x) = e ^{x}$ and prove that $\sum_{k=0}^{\infty} \frac{x^k}{k!}$ shares these properties.

$$ f \left( 0 \right) = 1 \\ \frac{d}{dx} f = f $$

If $\frac{d}{dx} y = y$ , then $y = Ce^x$. If $y(0) = Ce^0$ = $1$, then $C = 1$. Therefore $e^x$ is the only function with both these properties.

Now it's simple enough to show that $\sum_{k=0}^{\infty} \frac{x^k}{k!}$ has these same defining features.

Answer 4

Assume that we know nothing about the exponential function and the behavior of the limit of $(1+x/n)^n$ as $n\to\infty$. How would we establish the convergence and its equivalence to the power series $f(x)$ given by

$$ f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} \ ? $$

A rough idea may be to expand using binomial theorem and take limit termwise, but such manipulation needs some justification since interchangng limitig operators may be invalid in some cases. (If you know the dominated convergence theorem, then you can utilize it in a straightforward way. But this is like nuking a mosquito.) So here is one possible solution to ths technical issue:

To this end, we utilize the binomial theorem to expand

$$\left(1 + \frac{x}{n}\right)^n = \sum_{k=0}^{n} \binom{n}{k}\frac{x^k}{n^k} = \sum_{k=0}^{\infty} \frac{n(n-1)\cdots(n-k+1)}{n^k} \frac{x^k}{k!}. $$

The last equality holds because $ n(n-1)\cdots(n-k+1) = 0$ for $k > n$. Now fix $N$, and for $n > N$ we decompose the difference as

$$ \left| \left(1+\frac{x}{n}\right)^n - f(x) \right| \leq \sum_{k=0}^{N} \left| \frac{n(n-1)\cdots(n-k+1)}{n^k} - 1 \right| \frac{|x|^k}{k!} + 2 \sum_{k=N+1}^{\infty} \frac{|x|^k}{k!}. $$

Taking limsup as $n \to \infty$, we see that

$$ \limsup_{n\to\infty} \left| \left(1+\frac{x}{n}\right)^n - f(x) \right| \leq 2 \sum_{k=N+1}^{\infty} \frac{|x|^k}{k!}. \tag{1} $$

But the left-hand side does not depend on $N,$ so by taking $N\to\infty$, we find that the left-hand side of (1) is actually 0. This proves that

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

as desired. ////