Why does the sum of the reciprocals of factorials converge to $e$?

Question

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.

Answer 1

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

Answer 2

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 $\mathbb{e}$

Answer 3

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.
$$\exp(x)\exp(y)=\sum_{n=0}^{\infty}\frac{x^n}{n!}\sum_{m=0}^{\infty}\frac{y^m}{m!}\\ =\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{x^ny^m}{n!m!}\\ =\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{(m+n)!}{m!n!}\frac{x^ny^m}{(m+n)!}\\ \text{Let $k=n+m$. Then sum along diagonals of constant $k$.}\\ =\sum_{k=0}^{\infty}\frac1{k!}\sum_{n=0}^{k}{k\choose n}x^ny^{k-n}\\ =\sum_{k=0}^{\infty}\frac1{k!}(x+y)^k\\=\exp(x+y) $$

Answer 4

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

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

Answer 5

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}

Answer 6

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 $e$).

Answer 7

Since you are calling for intuition, a term by term version. Intuitively, the inverse factorial coefficient $\frac{1}{n!}$ is the most natural, as it yields "some kind of invariance to differentiation", since:

$$\left(\frac{x^n}{n!}\right)' = \left(n\frac{x^{n-1}}{n!}\right) = \frac{x^{n-1}}{(n-1)!} = \frac{x^{m}}{m!}$$ with $m= n-1$.

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 by reindexing: $$ f'(x) = \sum_{n=0}^\infty (n+1) a_{n+1}{x^{n}}\,,$$ 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 $$ f(1) = \sum_{n=0}^\infty \frac{1}{n!}=e\,.$$

Answer 8

Let $$I_n=\int_1^e\frac{(\ln x)^n}{x^2}dx.$$ Using integration by parts we have $$I_{n+1}=-\frac{1}{e}+(n+1)I_n.$$ Using induction we can prove that $$\frac{I_n}{n!}=1-\frac{1}{e}\sum_{k=0}^n \frac{1}{k!}$$ since $(I_n)_n$ is bounded, we let $n\to \infty$ we get $$\sum_{k=0}^\infty \frac{1}{k!}=e.$$

Answer 9

By definition
$$\frac{d}{dx}e^x = e^x$$ therefore $$\int_0^xe^t*1dt=e^x-1$$ we can use integration by parts which state that: $$\int{f(x)g(x)dx}=f(x)\int g(x) - f^`(x)\int\int g(x)+...$$ let f(x)= e^x, g(x) = 1 $$e^x\int1-e^x\int\int1+...=e^x-1$$ we simplify to get: $$\frac{1}{e^x}=1-\int1+\int\int1-\int\int\int1-...$$ we know that $\int\int..n..\int1dx=\frac{x^n}{n!}$ (basic integration rule)
therefore $$\frac{1}{e^x}=1-\frac{x^1}{1!}+\frac{x^2}{2!}-...$$

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let substitute x = -n, we know that (-x)^n = -x^n when n is odd $$e^n = 1+\frac{x^1}{1!}+\frac{x^2}{2!}+..$$ in fact, if we are to substitute x = ln(f(n)), n > 0 we get a nice series: $$f(n)=1+\frac{ln(f(n))^1}{1!}+\frac{ln(f(n))^2}{2!}+..$$