10
$\begingroup$

I've been asked by some schoolmates why we have $$ \sum_{n=0}^\infty \frac{1}{n!}=e.$$ I couldn't say much besides that the $\Gamma$ function, analytic continuation of the factorial, is defined with an integral involving $e$. Then I also know that actually $$ \sum_{n=0}^\infty \frac{x^n}{n!}=e^x.$$ Is there a reason for these facts?

P.S. I added the tag "intuition", please remove it if you think it is not pertinent.

$\endgroup$
  • 8
    $\begingroup$ How are you defining $e$? This, and its links, shows the usual ways of defining it are equivalent. $\endgroup$ – David Mitra Dec 6 '14 at 11:19
  • $\begingroup$ Well, I think it's rather $ \sum_{n=0}^\infty \frac{x^n}{n!}=e^x$, right ? $\endgroup$ – servabat Dec 6 '14 at 11:31
  • $\begingroup$ @servabat Oops, yeah, bad typo. Thanks for pointing it out. $\endgroup$ – Vincenzo Oliva Dec 6 '14 at 11:33
  • $\begingroup$ @DavidMitra We're using the limit one. Thanks for the link, that's useful. $\endgroup$ – Vincenzo Oliva Dec 6 '14 at 11:41
  • $\begingroup$ The case $x=1$ of Combinatorial proof answers this. This has been asked before. $\endgroup$ – robjohn Dec 6 '14 at 12:28
14
$\begingroup$

By definition, $$e=\lim_{n\to\infty}\left(1+\frac1n\right)^n.$$ Using the binomial theorem, the $k^{th}$ term of the development is $${\binom nk}\frac1{n^k}=\frac{n(n-1)(n-2)\dots(n-k+1)}{k!.n.n.n\dots n},$$ and $$\lim_{n\to\infty}{\binom nk}\frac1{n^k}=\frac1{k!}.$$

For example, $$\left(1+\frac1{1000}\right)^{1000}=\frac1{0!}+\frac1{1!}+\frac{0.999}{2!}+\frac{0.997002}{3!}+\frac{0.994010994}{4!}\dots$$

$\endgroup$
  • 4
    $\begingroup$ There are convergence problems here. That each term of a sum converges to a limit doesn't immediately tell you that the limit of the finite sums is the infinite sum of the limits. $\endgroup$ – Jack M Dec 6 '14 at 14:39
  • $\begingroup$ Not a proof, just intuition, as requested. $\endgroup$ – Yves Daoust Dec 6 '14 at 14:40
  • 1
    $\begingroup$ @Jack: Although, of course, there are many broad theorems which says this kind of argument works. $\endgroup$ – Hurkyl Dec 6 '14 at 14:45
  • $\begingroup$ I appreciated the other answers too, but yours really is the one that best gives me the reason I was thinking of. After all, I haven't studied yet Taylor series, so although I understood other arguments, this one regarding the definition I dealt with is to me the most natural. Thank you and the others! $\endgroup$ – Vincenzo Oliva Dec 6 '14 at 15:21
9
$\begingroup$

These two familiar sums are the Taylor series for $e^x$ about $0$. To get $e$ itself, you evaluate this series at $x=1.$

Derivation: The $n$th term of the Taylor series of a function $f$ about $a$ is

$$ \frac{f^{(n)}(a)}{n!} (x-a)^n.$$

But if $f(x) \triangleq e^x$, then $f'(x) = e^x$, and by an inductive argument, $f^{(n)}(x) = e^x$ for every positive integer $n.$ Taking the series about $a = 0,$ the $n$th term is

$$ \frac{f^{(n)}(a)}{n!} (x-a)^n = \frac{e^a}{n!} (x-a)^n = \frac{e^0}{n!} (x-0)^n = \frac{x^n}{n!}.$$

That is, the Taylor series of $e^x$ as a function of $x$ about $0$ is

$$ e^x = \sum_{n=0}^\infty \frac{x^n}{n!}, $$

and by setting $x=1$ we get

$$ e = \sum_{n=0}^\infty \frac{1}{n!}, $$

$\endgroup$
3
$\begingroup$

Since you are calling for intuition, a term by term version. You may know that, and start with, a fundamental property of the natural exponential: it is equal to its derivative.

So, suppose that
$$ f(x) = \sum_{n=0}^\infty a_n{x^n}\,,$$ is equal to its derivative. Then formally (I am skipping issues on convergence), $$ f'(x) = \sum_{n=1}^\infty n a_n{x^{n-1}}\,,$$ thus $$ f'(x) = \sum_{n=0}^\infty (n+1) a_{n+1}{x^{n}}\,,$$ then, one should have, term by term: $$ a_n= (n+1) a_{n+1}\,,$$ hence $$ a_{n+1} = \frac{a_n}{n+1} =\frac{a_0}{(n+1)!}\,.$$ Since the exponential is the reciprocal to the $\log$, you require that $f(0)=1$, hence $a_0=1$. So naturally,

$$ f(x) = \sum_{n=0}^\infty \frac{x^n}{n!}=e^x\,,$$ and $$ f(1) = \sum_{n=0}^\infty \frac{1}{n!}=e\,.$$

$\endgroup$
2
$\begingroup$

Here is one approach.

Let $b > 1$. If you compute the derivative of the function $b^x$, you find that the answer is just $b^x$ multiplied by an (annoying) constant.

There is a value of $b$ for which this constant is equal to $1$. That's nice! With this special value of $b$, the derivative of $b^x$ is just $b^x$, the same thing we started with. That's a very neat property for a function to have.

This special value of $b$ is $e = 2.718 \ldots$.

It is now easy to compute the Taylor series of the function $e^x$ (centered at $0$). We find that \begin{equation} \tag{$\spadesuit$} e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots. \end{equation} This comes directly from the Taylor series formula \begin{equation} f(x) = f(x_0) + f'(x_0)(x - x_0) + \frac{f''(x_0)}{2!}(x - x_0)^2 + \cdots. \end{equation}

Plugging $x = 1$ into ($\spadesuit$) yields \begin{equation} e = \sum_{n=0}^{\infty} \frac{1}{n!}. \end{equation}

$\endgroup$
1
$\begingroup$

Let $x \in \mathbb{R}^+$.

$f:x\longrightarrow \exp(x) \in C^\infty(\mathbb{R}, \mathbb{R})$, hence we can write $\forall n \in \mathbb{N}$ :

$$\left|f(x)-\sum_{k=0}^n f^{(k)}(0)\frac{x^k}{k!}\right| \leq \frac{|x|^{n+1}}{(n+1)!}I_{n+1} \text{ where } I_{n+1}=sup_{[0,x]}|f^{n+1}|$$

And $\forall t \in [O,x] f^{(n)}(t)=\exp(t)$ which means $I_{n+1}=\exp(x)$

But $\displaystyle\lim _{n\to \infty }\left(\frac{|x|^{n+1}}{(n+1)!}\exp(x)\right)=\:0$

So $\displaystyle\lim _{n\to \infty }\left(\sum_{k=0}^n\frac{x^{k}}{k!}\right)=\exp(x)$

Just take $x = 1$ for $\mathbb{e}$

$\endgroup$
1
$\begingroup$

When you multiply $\exp(x)$ by $\exp(y)$ by that definition, you get $\exp(x+y)$. That is one of the exponent laws, and is why $\exp(x)=e^x$ for some number $e$. Then $e^1=\exp(1)$ which is your sum.
$$\exp(x)\exp(y)=\sum_{n=0}^{\infty}\frac{x^n}{n!}\sum_{m=0}^{\infty}\frac{y^m}{m!}\\ =\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{x^ny^m}{n!m!}\\ =\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{(m+n)!}{m!n!}\frac{x^ny^m}{(m+n)!}\\ \text{Let $k=n+m$. Then sum along diagonals of constant $k$.}\\ =\sum_{k=0}^{\infty}\frac1{k!}\sum_{n=0}^{k}{k\choose n}x^ny^{k-n}\\ =\sum_{k=0}^{\infty}\frac1{k!}(x+y)^k\\=\exp(x+y) $$

$\endgroup$
1
$\begingroup$

Here is a complete rigorous proof, not using power series/exponential function. Suppose we are just learning about limits and series - we didn't learn about uniform convergence, etc.

Take this as the definition of $e$: $e = \lim_{n \to \infty} (1 + \frac 1 n)^n$. Then it is trivial to show that for any FIXED $N$, no matter how large (but FIXED), we also have $\lim_{n\to\infty}(1 + \frac 1 n)^{n+N} = e$ as well.

Let $S_n = \sum_{k=0}^n\frac 1 {k!}$. By the binomial theorem, $(1+\frac 1 n)^n \le S_n$ for all $n$, so we get that the sum of the infinite series is at least $e$ (comparison term by term: ${n \choose k} \cdot \frac 1 {n^k} \le \frac 1 {k!}$ for all $k$ from $0$ to $n$). This is the easy part.

For the other part, FIX a natural number N (however large, but FIXED). Again by the binomial theorem, and through the same kind of computation, we have $S_N \le (1+\frac 1 n)^{n+N}$. We only need the first $N+1$ terms of the expanded sum from the binomial theorem; $\frac 1 {k!} \le { n+N \choose k} \cdot \frac 1 {n^k}$. Now keep $N$ fixed and let $n \to \infty$; this shows that $S_N \le e$. Finally, since this is true for EVERY $N$, we get the opposite inequality: the sum of the series of inverses of factorials is $\le e$. By the way, this also shows the series converges (which we could see in other ways, but this by itself is a complete proof - we showed the partial sums are bounded by $e$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.