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

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.

• The right handside is the taylor expansion of the left handside.
– M. T
Aug 23, 2015 at 17:24
• For what it's worth, it's usually not the case that $\frac{n!}{(n-k)!n^k}=1$. For example, $\frac{2!}{(2-2)!2^2} = \frac{1}{2}$. But that's okay, because it's the limiting behaviour that matters, not the finite approximations. Aug 23, 2015 at 17:33
• Aug 23, 2015 at 18:50
• I think that the series above is the usual definition of exp. I know this is a bit pedantic, but still probably worth noting. Aug 23, 2015 at 19:04
• Aug 23, 2015 at 23:38

# 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.

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.

# 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 $$\bbox[5pt, border:2.5pt solid #FF0000]{\lim_{n\to \infty}S_n = \sum_{n=0}^\infty \frac{x^n}{n!}=e^{\,x}}$$

• How would you prove that the exponential function is indeed analytic? Aug 23, 2015 at 18:41
• @LeGrandDODOM That depends on which of several equivalent definitions of exponential function do you choose, and which domain do you consider. In this specific case $e^x$ is defined on reals, and we could simply show that the function is infinitely differentiable and that its Taylor series converges to the function. The former is obvious, the latter follows directly from the way we constructed our series. Our series happened to coincide with Taylor series for Exponent, so the function is indeed analytic. If we were unable to construct such series, we would not have analyticity of $e^x$.
Aug 23, 2015 at 19:05
• I want to prove that $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$ from $e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n$, so you don't answer to my question.
– idm
Aug 23, 2015 at 19:49
• @idm I am sorry my solution is unsatisfactory for you as it doesn't use $e^x=\lim_{n\to\infty}(1+x/n)^n$. However, I have to point out that you have not specified this requirement in your original post. The only explicit question you are asking is "How to prove that $e^x=\sum_{k=0}^\infty x^k/k!$?" The rest of your post looks like attempted solution; it is not clearly stated that the proof has to rely on a specific expression. Please edit your post in order to make clear all the requirement for the solution. Otherwise people may keep waisting time typing invalid solutions for you.
Aug 23, 2015 at 20:51
– idm
Aug 23, 2015 at 21:23

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 $n\rightarrow\infty$

• The sum of interest extends from $k=0$ to $k=n$. So, what happens to the limit (as $n\to \infty$) of the term in the post herein for $k=n$, $k=n-1$, $k=n-N$ for any fixed $N$? Aug 23, 2015 at 21:27
• As Euler put it, when $j$ is infinite, $\frac{j-1}j=1$. Euler played a lot with infinities. He literally meant that $j$ was an infinite number. I'm pretty sure he knew that that wasn't rigorous; I'm not sure he cared. This is the same man that wrote $\ln p=\dfrac{p^0-1}0$, an equation I find delightful. Aug 24, 2015 at 3:13
• (The "true" interpretation of that equality is $\displaystyle\ln p=\lim_{\epsilon\to0}\frac{p^\epsilon-1}\epsilon$, by the way.) Aug 24, 2015 at 3:14

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. ////

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.

• The trouble is to prove those properties from the definition as $\lim_{n \to \infty} (1 + x / n)^n$... Aug 23, 2015 at 23:15
• @vonbrand Well, evaluating $\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right) ^{n}$ at $x = 0$ is trivial. And, I believe it may be possible to say that $\frac{d}{dx} \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right) ^{n} = \lim_{n \to \infty} \frac{d}{dx} \left( 1 + \frac{x}{n} \right) ^{n}$. It is easy to show that this second expression is equal to the original limit. Is this correct? Aug 24, 2015 at 1:07