Approximate $\sqrt{e}$ by hand I have seen this question many times as an example of provoking creativity. I wonder how many ways there are to approximate $\sqrt{e}$ by hand as accurately as possible. 
The obvious way I can think of is to use Taylor expansion. 
Thanks 
 A: How accurately do you need it? One option is to use binomial expansion:
$$
e^{\frac{1}{2}} \approx \Big(1+\frac{1}{n}\Big)^{\frac{n}{2}}=\sum_{k=0}^{\frac{n}{2}}\binom{\frac{n}{2}}{k}\frac{1}{n^k}
$$
which you can make arbitrarily close to $e^{\frac{1}{2}}$ for various values of $n$.
A: I would use the fact that $e \approx 2.7182818284$ and use Wikipedia on computing square roots.  The digit by digit method will get you five decimals fairly quickly
A: As an alternative to series-based methods, there are differential equation based methods you can use.
If we recognize that $y=e^x$ is the solution to $y'=y$ with $y(0)=1$, and use Runge-Kutta with a small step size to approximate $y(1/2)$. 
In this case, with just one step (using $h=1/2$ in the link), we obtain $e^{1/2}\approx1.6484375\ldots$, compared to the actual value of $e^{1/2}=1.64872127\ldots$. 
With two steps, using $h=1/4$, we obtain $1.648699\ldots$.
With three steps, using $h=1/6$, we obtain $1.648716\ldots$.
With four steps, using $h=1/8$, we obtain $1.648716\ldots$.
With four steps, using $h=1/8$, we obtain $1.648719\ldots$.
In fairness, each "one" step in Runga-Kutta applied to this situation does require about seven multiplications. And since the step size needs to be decided from the start, you don't have the ability to refine your result further like you can with series by adding more terms. On the other hand a differential equation based method can give more accuracy in exchange for less computation in many cases.
A: And here is another answer. There is a known continued fraction expansion for $e^{1/n}$. Continued fraction sequences converge quickly (although with so many 1s, this particular continued fraction converges on the slower end of things). The downside is that you can't use the $n$th convergent to quickly find the $n+1$st convergent, so you have to make a choice right away how deep to go. As @spin notes in the comments, you can refine your convergent using the previous two convergents and the next integer in the continued fraction expression.
$$e^{1/2}=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{5+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{9+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{13+\cfrac{1}{1+\cfrac{1}{1+\cdots}}}}}}}}}}}}$$
A: Use the Power series to compute $x_n := \exp\left(-2^{-n}\right)$ for some $n \geqslant 1$ with high accuracy, and then compute
$$\sqrt{e} = \left(\frac{1}{x_n}\right)^{2^{n-1}}.$$
Using the negative exponent, and an exponent of smaller absolute gives you (much) faster convergence of the series, and the few operations of squaring and inverting don't lose much precision then. For e.g. $n = 3$, you get pretty good results for the first $10$ terms of the power series already.
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\root{\expo{}_{n}} = x_{n}}$. With Newton-Rapson: $\ds{x_{n + 1} = \half\pars{x_{n} + {e \over x_{n}}}\,,\quad x_{0} = 2}$

$\ds{\large\tt\mbox{With}\quad \color{#66f}{\Large n = 3}}$:
  \begin{align}&{\LARGE\sqrt{\expo{}}}\approx
\frac{1}{2} \left\{\frac{1}{2} \left[\frac{1}{2} \left(2+\frac{e}{2}\right)+\frac{2 e}{2+\frac{e}{2}}\right]+\frac{2 e}{\frac{1}{2} \left(2+\frac{e}{2}\right)+\frac{2 e}{2+\frac{e}{2}}}\right\} = x_{3}
\approx 1.648721295
\end{align}

$$
\mbox{Relative Error} = \verts{{x_{3} \over \root{\expo{}}} - 1}\times 100\ \%  =1.48\times 10^{-6}\ \%
$$

A: I found this series representation of $e$ on Wolfram Mathworld:
$$
e=\left(\sum_{k=0}^\infty\frac{4k+3}{2^{2k+1}(2k+1)!}\right)^2.
$$
Hence
$$
\sqrt{e}=\sum_{k=0}^\infty\frac{4k+3}{2^{2k+1}(2k+1)!}.
$$
Also from Maclaurin series for exponential function
$$
e^{\large\frac{1}{2}}=\sum_{n=0}^\infty\frac{1}{2^n n!}.
$$
A: The rapidly-converging series representation of $\sqrt{e}$ in Tunk-Fey's answer can be derived from simply expressing the Maclaurin series of $e^{x}$ as the sum of its even terms plus the sum of its odd terms.
$$ \begin{align} e^{x}&= \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} + \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}  = \sum_{n=0}^{\infty} \frac{x^{2n}(2n+1) + x^{2n+1}}{(2n+1)!} \\ &= \sum_{n=0}^{\infty} \frac{x^{2n}(2n+1+x)}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{x^{2n}(4n+2+2x)}{2(2n+1)!} \end{align}$$
A: If you apply the standard series expansion of $e^x$ to the case $x=-1/2$ and then find the reciprocal, it will converge faster than if you use $x=1/2$.
A: On a pocket calculator enter  $2048$, ${1\over x}$, $+$, $1$, $=$, $x^2$ ($10$ times).
A: You can use the following identities:


*

*$e=\lim_n(1+1/n)^n$,

*$e=\lim_n\frac{n}{^n\sqrt{n!}}$,
Put a large value of $n$ and it should do. Of course you have to square-root the results.
