How to prove $\sum_{n=0}^\infty \frac1{n!}=e\ $? How to prove $\displaystyle\sum_{n=0}^\infty \frac1{n!}=e\ $?
I thought about it but I could not find a proof. Please give me some hints?
 A: First look at the derivative of $2^x$.
$$\begin{align}
\frac{d}{dx} 2^x & = \lim_{\delta \rightarrow 0} \frac{2^{x+\delta} - 2^x}{\delta}
\\ & = 2^x \lim_{\delta \rightarrow 0} \frac {2^\delta - 1}{\delta}
\end{align}
$$
Let $f(x) = \lim_{\delta \rightarrow 0} \frac{x^\delta - 1}{\delta}$, which can be evaluated numerically.  $\frac{d}{dx} 2^x = 2^x f(2)$ and $f(2) \approx 0.693$.  Likewise $\frac{d}{dx} 3^x = 3^x f(3)$ and $f(3) \approx 1.08$.  What's the derivative of $f$?
$$\begin{align}
\frac{d}{dx} f(x) & = \lim_{\gamma \rightarrow 0} \frac{f(x+\gamma) - f(x)}{\gamma}
\\ &= \lim_{\gamma \rightarrow 0} \dfrac{ \lim_{\delta \rightarrow 0} \dfrac{(x + \gamma)^\delta - 1}{\delta} - \lim_{\delta \rightarrow 0} \dfrac{x^\delta -1}{\delta}}
{\gamma}
\\ &= \lim_{\gamma \rightarrow 0} \dfrac{ \lim_{\delta \rightarrow 0} \dfrac{(x + \gamma)^\delta - x^\delta}{\delta}}{\gamma}
\\ &= \lim_{\delta \rightarrow 0} \dfrac{ \lim_{\gamma \rightarrow 0} \dfrac{(x + \gamma)^\delta - x^\delta}{\gamma}}{\delta}
\\ &= \lim_{\delta \rightarrow 0} \dfrac{ \frac{d}{dx} x^\delta}{\delta} = \lim_{\delta \rightarrow 0} \dfrac{ \delta x^{\delta - 1}}{\delta} = \dfrac{1}{x}
\end{align}$$
This tells us that $f$ is mononically increasing between 2 and 3, so there must be a unique value of $x$ for which $f(x) = 1$.
$e$ is defined as the number for which $f(e) = 1$ and $\frac{d}{dx} e^x = e^x$.  Think about the Taylor series for $e^x = a_0 + a_1 x + a_2 x^2 + a_3 x^3 ... = \sum_{i=0}^{\infty} a_i x^i$.  
It has a value of 1 at $x = 0$, so $a_0 = 1$.  Since it is equal to its derivative, the derivatives of all the individual terms must be equal, i.e. $\forall i \ge 0: a_i x^i = (i+1) a_{i+1} x^i$.  $a_0 = 1$ and $a_i = (i+1)a_{i+1}$ and $\frac{a_i}{i+1} = a_{i+1}$.  This means $a_i = \frac1{i!}$ and $e^x = \sum_{i=0}^{\infty}\frac{x^i}{i!}$.
Evaluate the polynomial at x=1 and you get your expression for $e$.  
A: There are many ways to define $e,$ but somewhere along the line you have to define it before you can hope to prove it equal to some other quantity. If you define the function $e^{x}$ by the power series $\sum_{n=0}^{\infty} \frac{x^{n}}{n!},$ which converges for all real $x,$ then you just need to substitute $x =1$ to obtain what you want. I don't imagine this was the intention when the question was set. 
There are two other fairly common rigorous ways to define $e,$ but one is really covered by what is said already- that is, if we define the $\log$ function by $\log(x) = \int^{x}_{1} \frac{1}{t}dt$ for $x >0,$ then by the fundamental theorem of calculus, $\log$ is differentiable on $(0,\infty),$ and $\log^{\prime}(x) = \frac{1}{x}$ for all positive $x.$ Hence $\log$ is strictly increasing on $(0,\infty),$ and, in particular, is one-to-one. We define $\exp$ to be the inverse function of $\log,$ and we find easily that 
$\exp^{\prime} = \exp,$ from which it easily follows that the Maclaurin series for $\exp(x)$ is $\sum_{n=0}^{\infty} \frac{x^{n}}{n!},$ which we have dealt with.
The method defining $e$ which makes this problem most challenging is as $e = \lim_{n \to \infty} (1 + \frac{1}{n})^{n}.$ However, note that for every $n >0,$ we have 
$$(1 + \frac{1}{n})^{n} = \sum_{k=0}^{n} \frac{n!}{n^{k}k!(n-k)!} $$ 
which may be rewritten as 
$$\sum_{k=0}^{n} \frac{1}{k!}  \prod_{j=1}^{k-1} ( 1 - \frac{j}{n}).$$
This is always less than 
$$\sum_{k=0}^{\infty} \frac{1}{k!}$$
and greater than 
$$\sum_{k=0}^{n} \frac{1}{k!}  ( 1 - \frac{k}{n})^{k-1}.$$
Now fix a large integer $M.$ Then when $n >M,$ the last sum is greater than $$\sum_{k=0}^{M} \frac{1}{k!}  ( 1 - \frac{M}{n})^{M-1}.$$
As $n \to \infty,$ the last expression tends to $\sum_{k=0}^{M} \frac{1}{k!}.$
We have proved that 
$$(1+\frac{1}{n})^{n} \leq \sum_{k=0}^{\infty}\frac{1}{k!}$$
for all $n,$ and that for any fixed integer $M,$ we have 
$$\lim_{n \to \infty}(1+ \frac{1}{n})^{n} \geq \sum_{k=0}^{M} \frac{1}{k!}.$$
Hence 
$$\lim_{n \to \infty} (1 + \frac{1}{n})^{n} = \sum_{k=0}^{\infty} \frac{1}{k!}.$$
A: Use the taylor formula, assuming that for you $e = \exp(1)$ and $\exp' = \exp$.
$$
e - \sum_{k=1}^n \frac{1}{k!} =  \int_0^1 \frac{(1-t)^n}{n!} \exp(t) dt
$$
Now use $0\le \exp(t) \le \exp(1) =C $ on $(0,1)$ to get
$$
0\le e - \sum_{k=1}^n \frac{1}{k!} \le C\int_0^1 \frac{(1-t)^n}{n!}  dt
= \frac C{n+1}\to 0
$$ 
A: Another way to prove it is using differential equations. It's fairly straightforward to show that there exists a unique function $f(x)$ such that $f'(x) = f(x)$ and $f(0) = 1$, which can be proven by noting that $\frac{d}{dx}f^{-1}(x) = \frac1x$ and $f^{-1}(1) = 0$, which implies $f^{-1}(x) = \int_1^x\frac1t dt$. 
The next step is to show that there exists a constant $e$ such that $f(x) = e^x$. Let $e = f(1)$. Notice that $\frac{f(x + y)}{f(y)}$, seen as a function of $x$, solves the same initial value problem $f'(x) = f(x)$ and $f(0) = 1$. Uniqueness therefore implies $$
\frac{f(x+y)}{f(y)} = f(x)
$$ 
and hence $f(x+1) = f(1) f(x)$, and therefore $f(x) = f(1)^x = e^x$. Next, observe that the series $\sum_{n=0}^\infty \frac{x^n}{n!}$ converges absolutely for all $x$ which can be seen by the root test. Differentiating term by term we can see that this sum is its own derivative, and of course at $x= 0 $ it converges to $1$. Hence $e^x = f(x) = \sum_{n=0}^\infty \frac{x^n}{n!}$. The conclusion is $$
e = f(1) = \sum_{n=0}^\infty \frac1{n!}
$$
