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 $+\infty$ ?
Proof that $e=\sum\limits_{k=0}^{+\infty}\frac{1}{k!}$
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0@cardinal let $x$=0, g(0)=0^1/1!+0^2/2!+...=0, isn't it? – 2013-05-28
3 Answers
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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0@Michael Hardy : Thanks a lot Michael sir, for editing the post more neatly and I whole-heatedly appreciate your efforts. – 2012-01-15
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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0@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
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 $e<3$.