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

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.

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.