HMMT February

Algebra Difficulty 4.6 AIME Find the answer United States

Problem:
Define $\operatorname{sgn}(x)$ to be $1$ when $x$ is positive, $-1$ when $x$ is negative, and $0$ when $x$ is $0$ . Compute
$\sum_{n = 1}^{\infty}\frac{\operatorname{sgn}(\sin(2^{n}))}{2^{n}}.$
(The arguments to $\sin$ are in radians.)

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:
Note that each of the following is equivalent to the next.

$\cdot\ \operatorname{sgn}(\sin (2^{n})) = +1$

$\cdot\ 0 < 2^{n} \bmod 2\pi < \pi$

$\cdot\ 0 < \frac{2^{n}}{\pi} \bmod 2 < 1$

The $n$ th digit after the decimal point in the binary representation of $\frac{1}{\pi}$ is $0$ .

Similarly, $\operatorname{sgn}(\sin (2^{n})) = -1$ if and only if the $n$ th digit after the decimal point in the binary representation of $\frac{1}{\pi}$ is $1$ . In particular, if $a_{n}$ is the $n$ th digit, then $\operatorname{sgn}(\sin (2^{n})) = 1 - 2a_{n}$ .

Thus, the desired sum is
$\sum_{n = 1}^{\infty}\frac{\operatorname{sgn}(\sin(2^{n}))}{2^{n}} = \sum_{n = 1}^{\infty}\frac{1 - 2a_{n}}{2^{n}} = \left(\sum_{n = 1}^{\infty}\frac{1}{2^{n}}\right) - 2\left(\sum_{n = 1}^{\infty}\frac{a_{n}}{2^{n}}\right) = \left[1 - \frac{2}{\pi}\right].$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.