Brilli the bug

Let $x$ denote the horizontal coordinate and $y$ the vertical coordinate. Without loss of generality, we can assume that the first step is north (in the $y$ -direction), then left (in the $x$ -direction), then south, fourth right, and so on ad infinitum. In this case, the final $x$ -displacement is $\begin{array}{rl} \text{x-displacement} &= \sum\limits_{n=0}^{\infty} \dfrac{2(-1)^{n}}{(2n + 1)^2}\\ &= 2\sum\limits_{n=0}^{\infty} \dfrac{(-1)^{n}}{(2n + 1)^2}\\ &= 2G \end{array}$ whereas the final $y$ -displacement is $\begin{array}{rl} \text{y-displacement} &= \sum\limits_{n=0}^{\infty} \dfrac{2(-1)^{n}}{(2n + 2)^2}\\ &= \dfrac{1}{2}\sum\limits_{n=0}^{\infty} \dfrac{(-1)^n}{(n + 1)^2}\\ &= \dfrac{1}{2} \left(\dfrac{\pi^2}{12}\right)\\ &= \dfrac{\pi^2}{24} \end{array}$ So the final displacement after infinite steps is $\begin{array}{rl} \sqrt{\left(\text{x-displacement}\right)^2 + \left(\text{y-displacement}\right)^2} &= \sqrt{\left(2G\right)^2 + \left(\dfrac{\pi^2}{24}\right)^2}\\ &= \dfrac{1}{24}\sqrt{2304G^2 + \pi^4} \end{array}$ where $\alpha = 24, \beta = 2304, \gamma = 2, \delta = 6$ . Thus, $\alpha + \beta + \gamma + \delta = \boxed{2334}$

Consider the complex number $z = x+iy$ , then $|z|$ is the displacement of Brilli from the origin. We note that $z_0 = \dfrac 1{(0+1)^2} = \dfrac 2{1^2}$ , $z_1 = \dfrac 2{1^2} + \dfrac {2i}{2^2}$ , $z_2 = \dfrac 2{1^2} + \dfrac {2i}{2^2} + \dfrac {2i^2}{3^2}$ and so on, implying:

$\begin{aligned} z_\infty & = \frac 2{1^2} + \frac {2i}{2^2} + \frac {2i^2}{3^2} + \frac {2i^3}{4^2} + \frac {2i^4}{5^2} + \frac {2i^5}{6^2} + \cdots \\ & = \frac 2i \left(\frac i{1^2} + \frac {i^2}{2^2} + \frac {i^3}{3^2} + \frac {i^4}{4^2} + \frac {i^5}{5^2} + \frac {i^6}{6^2} + \cdots \right) \\ & = \frac 2i \left(\frac i{1^2} - \frac 1{2^2} - \frac i{3^2} + \frac 1{4^2} + \frac i{5^2} - \frac 1{6^2} - \cdots \right) \\ & = 2 \left(\frac 1{1^2} + \frac i{2^2} - \frac 1{3^2} - \frac i{4^2} + \frac 1{5^2} + \frac i{6^2} + \cdots \right) \\ & = 2 {\color{#3D99F6}\left(\frac 1{1^2} - \frac 1{3^2} + \frac 1{5^2} - \cdots \right)} + 2i \left(\frac 1{2^2} - \frac 1{4^2} + \frac 1{6^2} + \cdots \right) \\ & = 2 {\color{#3D99F6}G} + \frac {2i}{2^2} \left(\frac 1{1^2} - \frac 1{2^2} + \frac 1{3^2} + \cdots \right) & \small \color{#3D99F6} \text{where }G \text{ is the Catalan's constant.} \\ & = 2G + \frac i2 \left(\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \cdots - 2 \left(\frac 1{2^2} + \frac 1{4^2} + \frac 1{6^2} + \cdots \right) \right) \\ & = 2G + \frac i2 \left(\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \cdots - \frac 2{2^2} \left(\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \cdots \right) \right) \\ & = 2G + \frac i4 \color{#3D99F6}\left(\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \cdots \right) \\ & = 2G + \frac i4 \cdot \color{#3D99F6} \zeta (2) & \small \color{#3D99F6} \text{where }\zeta (\cdot) \text{ is the Riemann zeta function.} \\ & = 2G + \frac i4 \cdot \color{#3D99F6} \frac {\pi^2}6 \\ & = 2G + i\frac {\pi^2}{24} \end{aligned}$

$\implies \left|z_\infty \right| = \sqrt{4G^2 + \dfrac {\pi^4}{24^2}} = \dfrac 1{24}\sqrt {4\cdot 24^2 G^2 + \pi^4}$

$\implies \alpha + \beta + \gamma + \delta = 24 + 4 \cdot 24^2 + 2+4 = \boxed{2334}$

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References:

• Catalan's constant
• Riemann zeta function