# Proof that $e=\sum\limits_{k=0}^{+\infty}\frac{1}{k!}$

How can it be proved that the Euler constant equals the limit of the sum of all $\frac{1}{k!}$ when $k$ goes from $0$ to $+\infty$ ?

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what is your definition of e? – m_t_ Jan 14 '12 at 15:15
and exp is... ? – Dustan Levenstein Jan 14 '12 at 15:20
no, we need a definition. Some people define exp as $\displaystyle\sum_{k=0}^\infty \frac{x^k}{k!}$ – Dustan Levenstein Jan 14 '12 at 15:23
If you grant that $\exp$ satisfies $f'=f$, try a MacLaurin expansion. – Neal Jan 14 '12 at 15:26
@Skydreamer that differential equation has lots of solutions. You have to impose y(0)=1. If you already believe that that DE has a unique solution, just verify that the power series defined by Dustan earlier solves it (differentiate term by term), then plug in x=1. – m_t_ Jan 14 '12 at 15:27

But I encountered the same doubt when I was reading the " Synopsis of elementary results in mathematics ", I convinced myself with this two facts ( I don't know whether they are true or not, that should be decided by Mr.Srivatsan ) .

The function $e^x$ has derivative equal to itself. Then the Maclaurin series for any function which can be differentiated as many times as you like is

$$f(x) =\large \frac{f(0)}{0!} + f^\prime(0)\cdot\large \frac{x}{1!} + f^{\prime\prime}(0)\cdot\large \frac{x^2}{2!} + f^{\prime\prime\prime}(0).\frac{x^3}{3!} + \cdots$$

For $f(x) = e^x$, you have

$e^x = f(x) = f^\prime(x) = f^{\prime\prime}(x) = f^{\prime\prime\prime}(x) = \cdots 1 = f(0) = f^\prime(0) = f^{\prime\prime}(0) = f^{\prime\prime\prime}(0) = \cdots$

and the Maclaurin series for $e^x$ is then

$$e^x =\large 1 + \frac{x}{1!} + \frac{x^2}{2!} +\frac{ x^3}{3!} + \frac{x^4}{4!} + \cdots$$

Now set $x = 1$, and you get the series about which you asked.

Another version:

The definition of $e$ is

$$e = \lim_{n\to \infty}(1+1/n)^n$$

Consider the binomial expansion for$n = 1, 2, 3, 4, 5, \ldots$

$$(1+1/n)^n = \sum^n_{i=0}C(n,i) (1/n)^i$$

For $i = 0, 1, 2, 3, \ldots$ one has

$$C(n,i)(1/n)^i = \rm{ \large \frac{n!}{(n-i)!i!n^i}}$$ $$= (1)(1-1/n)(1-2/n)\cdots (1-[i-1]/n)/i!$$

whose limit as n grows without bound is $\large\frac{1}{i!}$ . Then

$$\lim_{ n\to \infty} (1+1/n)^n = \lim_{ n\to \infty} \sum^n_{i=0} C(n,i) (1/n)^i$$ $$= \sum^\infty_{i=0} \lim_{n\to \infty} C(n,i)(1/n)^i$$

$$e = \sum^{\infty}_{i=0} 1/i!$$

Hence the result.

( Credits of editing goes to Mr.Srivatsan , as he taught me to use ' instead of \prime and many more things which made my answer appear more neatly, and also for Mr.Michael Hardy, for editing the answer which now appears more neatly ).

Thank you.

Yours truly,

Iyengar.

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+1 because of the second proof it is really beautiful and view from another direction. – speedyGonzales Jan 14 '12 at 20:11
As mentioned in the other answer based on the binomial expansion of $(1+1/n)^n$, one should add an argument justifying the $\lim\sum=\sum\lim$ step, to transform the part called Another version into a full proof. – Did Jan 14 '12 at 22:52
The first version is not a complete proof. It is false that a "function which can be differentiated as many times as you like" equals its Maclaurin series. Some times it does and some times it does not. (When it does, we say that the function is analytic.) See the references posted in comments at mathoverflow.net/questions/81613/… by Dave Renfro. – Andrés Caicedo Jan 15 '12 at 0:04
The issue is, one has to show that 1. The Maclaurin series converges and, if it does, that 2. It converges to the right value. We check that this is the case for $e^x$, but it is the key part of the proof. – Andrés Caicedo Jan 15 '12 at 0:06
@Michael Hardy : Thanks a lot Michael sir, for editing the post more neatly and I whole-heatedly appreciate your efforts. – Iyengar Jan 15 '12 at 3:17

I'll assume $e=\lim_{n\to\infty}(1+1/n)^n$. Here is a heuristic argument that can be made rigorous. Apply the binomial theorem to $(1+1/n)^n$ to get $$(1+1/n)^n=\sum_{k=0}^n \binom{n}{k}n^{-k}=1+n/n+\frac{n(n-1)}{2n^2}+\cdots$$ This is approximately $1+1+\frac{1}{2}+\frac{1}{3!}+\cdots.$ Taking the limit as $n$ goes to infinity, we get $e=\sum_{k=0}^\infty \frac{1}{k!}$.

I've made it a community wiki in case anyone wants to supply some of the missing details to make it fully rigorous.

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Thank you for the answer ! – Skydreamer Jan 14 '12 at 16:57
@Skydreamer : I have been editing the answer in TeX and another person posted the same answer before me, its my bad luck, all my strain gone in vain. Anyway you got the answer, thats good, thank you. – Iyengar Jan 14 '12 at 17:07
I gave you a +1 but I won't unaccept an answer, that's not a nice behavior. Thank you anyway ! – Skydreamer Jan 14 '12 at 17:23
@iyengar: This happens frequently. Two people will be putting in their answer at the same time. It's happened to me on many occasions. One strategy is to post a quick answer so people can see you're working on the problem, and then take the time to edit the answer into a more detailed form afterwards. (I learned this strategy from Bill Dubuque.) – Grumpy Parsnip Jan 14 '12 at 17:45
@Skydreamer: since I made this a CW, I don't get any points anyway, so there's no problem switching the accepted answer to iyengar's more complete answer. Some people get irritated about such things, but I'm not one of them. Cheers! – Grumpy Parsnip Jan 14 '12 at 17:55

You have to prove that the sequence of partial sums of the series converges. But for all $x$ , $e^x=1+x+....+x^n/n!+r(x)$ where $r(x)$ is the rest of order $n$. Prove that for $x=1$ the sequence $r(x)$ converges to zero. You can use the formula of Lagrange, and use $e<3$.

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