# Summing a geometric series: Secrets of the Mathematical Ninja

“I don’t know if you were actually working stuff out there, or if you just muttered at random for a bit and guessed,” said the Mathematical Ninja’s cheeky student.

Stung, the Mathematical Ninja was forced to explain how he figured out $\left(1 - \frac{1}{8}\right)^{19}$.

It’s a simple enough trick that he’s explained many times before: it involves knowing that $\left(1 + \frac{1}{k}\right)^{k} \simeq e.$ In fact, $\left(1 - \frac{1}{8}\right)^{8}$ is about six or seven percent away from $1/e$. As a rule of thumb, the percentage error is about $\frac{50}{n}$, and the fraction is lower than $1/e$.

That means, $\left(1 - \frac{1}{8}\right)^{19}$ is roughly $\left(\frac{1}{e} - 7\% \right)^{19/8}$. You need to multiply the error by the power ($\frac{50\times19}{8\times 8} \simeq \frac{1000}{64}$ - so that’ll make it pretty much $e^{-2.4} - 16%$).

But what’s $e^{-2.4}$? Well, any fule no that since $\ln(10) \simeq 2.3$ and $\ln(12) = 2\ln(2) + \ln(3) \simeq 1.4 + 1.1 = 2.5$, you can surmise that $e^{-2.4} \simeq \frac{1}{11}$.

That’s a tenth, less 10% (give or take), less another 16% - somewhere in the upper 0.07s. The correct answer? 0.079.

Muttering randomly, indeed.

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