I can't have anymore Pi-Pies...

Calculus Level pending

Suppose a Pie is a perfect circle with a radius of π \pi . Suppose I I is the area of the pie, and P P is the circumference of the pie, calculate the infinite series P I + P 2 I 2 + P 3 I 3 + . . . \frac { P }{ I } \quad +\quad \frac { { P }^{ 2 } }{ { I }^{ 2 } } \quad +\quad \frac { { P }^{ 3 } }{ { I }^{ 3 } } \quad +\quad ...

π π { \pi }^{ \pi } 2 π − 2 \frac { 2 }{ \pi \quad -\quad 2 } 1 π − 1 \frac { 1 }{ \pi \quad -\quad 1 } π π \sqrt [ \pi ]{ \pi } 1 π \frac { 1 }{ \pi } π \pi ∞ \infty

Abyoso Hapsoro by Abyoso Hapsoro , 22, Indonesia

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1 solution

Théo Leblanc
Apr 4, 2019

For a circle / disk:

P = 2 π × R P=2\pi\times R

I = π × R 2 I=\pi\times R^2

Thus P I = 2 R \frac{P}{I}=\frac{2}{R}

With R R the radius.

In our case, R = π R=\pi , so P I = 2 π \frac{P}{I}=\frac{2}{\pi}

Using the fact that ∀ z ∈ C , ∣ z ∣ < 1 , ∑ n = 0 ∞ z n = 1 1 − z \forall z \in \mathbb{C}, \ |z| <1, \ \displaystyle\sum_{n=0}^{\infty}z^n=\frac{1}{1-z}

∣ 2 π ∣ < 1 |\frac{2}{π}|<1 so the answer is 2 π × 1 1 − 2 π = 2 π − 2 \frac{2}{π} \times \frac{1}{1-\frac{2}{π}}=\boxed{\frac{2}{π-2}}

By the way, because we are summing quantities with m − 1 m^{-1} dimension at different exponents, it is not homogeneous (in a physical sense) so depending on the units system we use (meters, centimeters or other), the series may diverge or converge....

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