How to estimate the value of $e$. I am currently studying how to estimate $e$. To solve this problem I use these methods discuss below:
Method 1:
We know that $e^x = 1 + \dfrac{x}{1!} + \dfrac{x}{2!}+ \cdots $
So if we consider a large $n$ we can estimate $e$ as $1 + \dfrac{1}{1!} + \dfrac{1}{2!}+ \cdots + \dfrac{1}{n!}$
Method 2:
Another way is I consider $I=\displaystyle\int_1^{2} \dfrac1x \,dx =\log_{e} 2$. Now I can compute numerically the integral $I$. So from that I estimate $e$ as $e^I = 2 \implies e= 2^\frac1I$
I believe there is many other way to estimate the value of $e$. If you know please discuss here. 
 A: In $2004$, using series compression, Brothers proposed $$e=\sum_{n=0}^{\infty}\frac{2n+2}{(2n+1)!}$$ which converges extremely fast. 
Using $10$ terms, you get $$e\approx \frac{69439789852104840011}{25545471085854720000}\approx 2.71828182845905$$ which has an error of $9 \times 10^{-22}$ while, for the same number of terms, the classical expansion would give $$e\approx \frac{9864101}{3628800}\approx2.71828180114638$$ which has an error of $2.7 \times 10^{-8}$.
Using $20$ terms for the summation gives more than $50$ accurate digits.
In the same paper, Brothers also proposed $$e=\sum_{n=0}^{\infty}\frac{3-4n^2}{(2n+1)!}$$  $$e=\sum_{n=0}^{\infty}\frac{9n^2+1}{(3n)!}$$ This last one leads to $78$ correct digits after $20$ terms.
A: There is another well known formula:
$$e^x =\lim_{n\to \infty} (1+\frac x n)^n$$
Just plug in a $n$ great enough for an approximation.
If $n$ is a power of two $n = 2^k$ then you can calculate it very efficiently by initializing $y := 1+x/n$ Then repeating following $k$ times: $y := y\cdot y$.
There are other limits:
$$ e = \lim_{n\to\infty} \frac{n}{\sqrt[n]{n!}}$$
Other approaches could also be found using continued fraction expansions, e.g. Eulers identity:
\begin{align}
e &= [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1,\dots] \\
  &= 2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{4+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{6+\dotsb}}}}}}}}
\end{align}
A: You can use the continued fraction
$$
e = 2 + \cfrac{1}{\color{red}1+\cfrac{1}{\color{red}2+\cfrac{1}{\color{red}1 + \cfrac{1}{\color{red}1+\cfrac{1}{\color{red}4 + \cdots}}}}}
$$
where the pattern of the red numbers goes $1,2,1,1,4,1,1,6,1,1,8,1,1,10,1\ldots$ and so on in the obvious way. Terminate it wherever you want, and you cet in some sense the best rational number approximation you can get.
While I haven't looked into it, it is entirely possible that this is equivalent to your method 1.
A: What about just using the Taylor series for $e^{-1}$? This gives, after a little rearrangement : 
$\frac{1}{e} = \sum_{n=1}^{\infty} \frac{ 2n}{(2n+1)!},$ which converges pretty fast.
A: Using a better converging continued fraction from e.g. Wikipedia
$$e = 1 + \cfrac{2}{1+\cfrac{1}{6+\cfrac{1}{10 + \cfrac{1}{14+\cfrac{1}{18 + \cfrac{1}{22+\cfrac{1}{26+\cdots}}}}}}}
$$
you get the approximation
$$
e \approx \frac{28245729}{10391023} 
$$
which has an error of $6.16\cdot 10^{-16}.$
A: $$ e=\lim_{n \to \infty }(1+\frac{1}{n})^n=\binom{n}{0}1^n\frac{1}{n}^0+\binom{n}{1}1^{n-1}\frac{1}{n}^1+\binom{n}{2}1^{n-2}\frac{1}{n}^2+\binom{n}{3}1^{n-3}\frac{1}{n}^3+...\\=1+n\frac{1}{n}+\frac{n(n-1)}{2}\frac{1}{n^2}+\frac{n(n-1)(n-3)}{3!}\frac{1}{n^3}+...=\\1+1+\frac{n-1}{2n}+\frac{(n-1)(n-2)}{6n^2}+\frac{(n-1)(n-2)(n-2)}{24n^3}+...$$
A: $e$ is a root of the function
$$ f(x) = (\ln x) - 1 $$
whose solution we can estimate using Newton's algorithm, by picking a starting value $x_0$ and setting
$$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n (2 - \ln x_n) $$
Of course, to do this, we need to be able to estimate the natural logarithm. (for computer calculation, I understand that there are clever methods to compute logarithms efficiently, and this iteration is one of the main methods for actually computing $\exp(x)$)

Another approach inspired by computer algorithms is that
$$ e^1 = (e^{1 / 2^k})^{2^k} $$
Since $1/2^k$ is very small, we can use other methods to quickly get a very good estimate for $e^{1/2^k}$. Then we square that $k$ times.
