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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?2012-01-14
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    This is done in Baby Rudin (sequences and series chapter), and it's kind of a sticky (albeit simple) proof.2012-01-14
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    @mt_ $e=\exp(1)$. Alex I'll take a look at it.2012-01-14
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    and exp is... ?2012-01-14
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    The exponential function.2012-01-14
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    no, we need a definition. Some people define exp as $\displaystyle\sum_{k=0}^\infty \frac{x^k}{k!}$2012-01-14
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    The only solution of the differential equation $y'=y$ where $y'$ is the derivative of $y$2012-01-14
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    If you grant that $\exp$ satisfies $f'=f$, try a MacLaurin expansion.2012-01-14
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    @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=1.2012-01-14
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    @mt_ Of course, I forgot it. Neal, I'll read about this now, thank you !2012-01-14
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    If (as in one of the usual definitions), $e$ is defined as the limit as $n$ goes to infinity of $(1+1/n)^n$, there is a verification of what you want by expanding using the Binomial Theorem, and making some estimates. I expect that has been done (and more than once!) on **this** site, but my searching skills are limited.2012-01-14
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    This is done, e.g., in Lang's *Undergraduate Analysis*. The proof may not be all that enlightening, though. Essentially, one starts with the definition of the *function* $\exp(x)$ as the unique function such that $f'(x) = f(x)$ for all $x$ and $f(0) = 1$. (This is easy.) Then, once one develops standard facts of power series, one shows that the function $g(x) = \sum_{n=1}^\infty x^n/n!$ satisfies the above properties. So, $g \equiv f$. Now, plug in $x=1$.2012-01-14
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    @cardinal let x=0, g(0)=0^1/1!+0^2/2!+...=0, isn't it?2013-05-28

3 Answers 3

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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.2012-01-14
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    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.2012-01-14
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    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 http://mathoverflow.net/questions/81613/non-analytic-function-with-convergent-taylor-series-everywhere by Dave Renfro.2012-01-15
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    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.2012-01-15
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    @Michael Hardy : Thanks a lot Michael sir, for editing the post more neatly and I whole-heatedly appreciate your efforts.2012-01-15
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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 !2012-01-14
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    @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.2012-01-14
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    I gave you a +1 but I won't unaccept an answer, that's not a nice behavior. Thank you anyway !2012-01-14
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    Not a nice behavior, I behaved properly , I told frankly that my strain was wasted, If I had known that another person is posting the same answer, I might not have started TeXing all the answer, as the both answers are same. Why are you thinking that I mis-behaved. Can I ask you for Justification of word " Mis-behavior " ? @Skydreamer2012-01-14
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    I am not thinking you misbehaved. The last time I gave the points to another person after I accepted an answer, I got a rocket so I don't know if I should do it or not.2012-01-14
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    @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.)2012-01-14
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    @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!2012-01-14
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    Okay @JimConant, I give him the points !2012-01-14
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    @JimConant : Sir, Your strategy is very nice, in fact I am much happy to learn about your procedure, Thanks a lot !! . But I don't have desire to earn reputation, but as you know Human's efforts are fulfilled when they are appreciated. Without appreciation no human will be happy, that too, I am a privately learning self-taught mathematician, I am very new to this TeX, I am following guidelines given by community, so it takes a lot of time for editing, ( and in this case it took me 30 minutes ), so I was suddenly disappointed after making up long answer, you have answered the same. ..Contd2012-01-15
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    @JimConant : Contd.... , So I told him that if I had known previously that you are posting the same answer, I might not have started answering it, as I know that you are far better to me in knowledge as well as experience. And I stress again , I am not craving for reputation sir, but only worrying about the wasted efforts, I hope you understood. Once again Thanks a ton !!, and I am extremely sorry if I have made you hurt in any situation .2012-01-15
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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 $e<3$.