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

Solution 1:

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 $$ e^{\large\frac{1}{2}}=\sum_{n=0}^\infty\frac{1}{2^n n!}. $$

Solution 2:

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.

$$ \begin{align} e^{x}&= \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} + \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{x^{2n}(2n+1) + x^{2n+1}}{(2n+1)!} \\ &= \sum_{n=0}^{\infty} \frac{x^{2n}(2n+1+x)}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{x^{2n}(4n+2+2x)}{2(2n+1)!} \end{align}$$

Solution 3:

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 $x=1/2$.

Solution 4:

On a pocket calculator enter $2048$, ${1\over x}$, $+$, $1$, $=$, $x^2$ ($10$ times).

Solution 5:

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 $n$.