Two Infinite Series Sums from Regular Polygons (visual proof) 

I don't think these 2 are the same. But maybe that's because I don't know what an Apeirogon is.

If you could see infinity, there would be a slight difference

OK, I checked the definition of an Apeirogon and I must say, accprding to its definition, it is not at all related to circles And @daniel_77 what does "if you could see infinity" mean? And how does it imply that an Apeirogon and a circle are different. They are not similar at all to start with.

A (degenerate) apeirogon is a circle in the context of projective geometry, but in affine geometry, where these types of mathematics are being done, they are completely different shapes.

@@angelmendez-rivera351 WOW! That's totally unexpected! Care to tell us what the definition in both contexts and explain the difference please?

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For sure :) still kinda telescoping.

Thank you! 😊

​@@MathVisualProofs what's the proof a circle has 0 sides

Thanks!

Thank you!!

In this case, the circle should actually be infinitely large. It would just be a horizontal line

@@duane6386 ah, right. I had in mind an inscribing circle, which this is not.

​@@wieneryron Area of a regular polygon of 𝒏 ∈ ℕ, 𝒏 ≥ 3 sides and side of length 𝐚 ∈ ℝ₊*. (Read and write it in a paper at the same time) Let 𝐚ₖ, 𝒌 ∈ ℕ* be the 𝒌-th segment of the polygon. Slice the polygon in 𝒏 equilateral triangles in which each base is one of the 𝐚ₖ segments. Let 𝐀 be the center of the polygon and let 𝐁ₖ,𝐂ₖ be the vertexes of each 𝐚ₖ. In the equilateral triangle 𝚫𝐀𝐁₁𝐂₁, that is congruent to the others, we have that the internal angle of the vertex 𝐀 is equal to a full cicle (2𝛑) divided by the quantity of sides: 𝐀𝐧𝐠(𝐀) = 2𝛑/𝒏. Taking the segment 𝐇, that represents the height of the triangle and whose lenght is equal to 𝒉 ∈ ℝ₊*, whose vertexes are 𝐀 and 𝐃₁ such that 𝐃₁ divides the segment 𝐚₁ in two congruent segments of length 𝐚/2, and that divides the internal angle of the center in a half, we have that, in the triangle 𝚫𝐀𝐁₁𝐃₁, 𝒉 = 𝐚𝐂𝐨𝐭(𝛑/𝒏)/2. It's trivial that the area of a triangle is equal to the half of the product between the lenght of the base and the lenght of height in respect to such base. The area of 𝚫𝐀𝐁₁𝐃₁ is a half of the area of the triangle 𝚫𝐀𝐁₁𝐂₁. The are of the polygon is equal to 𝒏 multiplied by the are of the triangle 𝚫𝐀𝐁₁𝐂₁. Therefore, the area of the polygon is equal to 𝒏•2• (1/2)•(𝐚/2)•[𝐚𝐂𝐨𝐭(𝛑/𝒏)/2] = 𝐚²𝐂𝐨𝐭(𝛑/𝒏)/4. Determining the limit of the are of the polygon when 𝒏 goes to infinity: 𝐋𝐢𝐦 [𝐚²𝐂𝐨𝐭(𝛑/𝒏)/4] = 𝒏 → ∞ (𝐚²/4) 𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] = 𝒏 → ∞ 𝛑 is positive, so 𝛑/𝒏 when 𝒏 goes to infinity is also positive. When 𝒏 goes to infinity, 1/ 𝒏 goes to 0 by positive values. When 1/ 𝒏 goes to 0 by positive values, 𝐂𝐨𝐭(1/ 𝒏) goes to infinity. Therefore: 𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] = ∞ 𝒏 → ∞ 𝐚²/4 is always positive. Therefore: (𝐚²/4) 𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] = ∞ 𝒏 → ∞ Conclusion: the "circle" in as big as you want it to be.

​​@@wieneryron Area of a regular polygon of 𝒏 ∈ ℕ, 𝒏 ≥ 3 sides and side of length 𝐚 ∈ ℝ₊*. (Read and write it in a paper at the same time) Let 𝐚ₖ, 𝒌 ∈ ℕ* be the 𝒌-th segment of the polygon. Slice the polygon in 𝒏 equilateral triangles in which each base is one of the 𝐚ₖ segments. Let 𝐀 be the center of the polygon and let 𝐁ₖ,𝐂ₖ be the vertexes of each 𝐚ₖ. In the equilateral triangle 𝚫𝐀𝐁₁𝐂₁, that is congruent to the others, we have that the internal angle of the vertex 𝐀 is equal to a full cicle (2𝛑) divided by the quantity of sides: 𝐀𝐧𝐠(𝐀) = 2𝛑/𝒏. Taking the segment 𝐇, that represents the height of the triangle and whose lenght is equal to 𝒉 ∈ ℝ₊*, whose vertexes are 𝐀 and 𝐃₁ such that 𝐃₁ divides the segment 𝐚₁ in two congruent segments of length 𝐚/2, and that divides the internal angle of the center in a half, we have that, in the triangle 𝚫𝐀𝐁₁𝐃₁, 𝒉 = 𝐚𝐂𝐨𝐭(𝛑/𝒏)/2. It's trivial that the area of a triangle is equal to the half of the product between the lenght of the base and the lenght of height in respect to such base. The area of the polygon is equal to 𝒏 multiplied by the area of the triangle 𝚫𝐀𝐁₁𝐂₁. Therefore, the area of the polygon is equal to 𝒏•(1/2)•𝐚•[𝐚𝐂𝐨𝐭(𝛑/𝒏)/2] = 𝐚²𝐂𝐨𝐭(𝛑/𝒏)/4. Determining the limit of the are of the polygon when 𝒏 goes to infinity: 𝐋𝐢𝐦 [𝐚²𝐂𝐨𝐭(𝛑/𝒏)/4] = 𝒏 → ∞ (𝐚²/4) 𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] = 𝒏 → ∞ 𝛑 is positive, so 𝛑/𝒏 when 𝒏 goes to infinity is also positive. When 𝒏 goes to infinity, 1/ 𝒏 goes to 0 by positive values. When 1/ 𝒏 goes to 0 by positive values, 𝐂𝐨𝐭(1/ 𝒏) goes to infinity. Therefore: 𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] = ∞ 𝒏 → ∞ 𝐚²/4 is always positive. Therefore: (𝐚²/4) 𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] = ∞ 𝒏 → ∞ Conclusion: the "circle" is as big as you want it to be.

​​​@@wieneryron Area of a regular polygon of 𝒏 ∈ ℕ, 𝒏 ≥ 3 sides and side of length 𝐚 ∈ ℝ₊*. (Read and write it in a paper at the same time) Let 𝐚ₖ, 𝒌 ∈ ℕ* be the 𝒌-th segment of the polygon. Slice the polygon in 𝒏 equilateral triangles in which each base is one of the 𝐚ₖ segments. Let 𝐀 be the center of the polygon and let 𝐁ₖ,𝐂ₖ be the vertexes of each 𝐚ₖ. In the equilateral triangle 𝚫𝐀𝐁₁𝐂₁, that is congruent to the others, we have that the internal angle of the vertex 𝐀 is equal to a full cicle (2𝛑) divided by the quantity of sides: 𝐀𝐧𝐠(𝐀) = 2𝛑/𝒏. Taking the segment 𝐇, that represents the height of the triangle and whose lenght is equal to 𝒉 ∈ ℝ₊*, whose vertexes are 𝐀 and 𝐃₁ such that 𝐃₁ divides the segment 𝐚₁ in two congruent segments of length 𝐚/2, and that divides the internal angle of the center in a half, we have that, in the triangle 𝚫𝐀𝐁₁𝐃₁, 𝒉 = 𝐚𝐂𝐨𝐭(𝛑/𝒏)/2. It's trivial that the area of a triangle is equal to the half of the product between the lenght of the base and the lenght of height in respect to such base. The are of the polygon is equal to 𝒏 multiplied by the are of the triangle 𝚫𝐀𝐁₁𝐂₁. Therefore, the area of the polygon is equal to 𝒏•(1/2)•𝐚•[𝐚𝐂𝐨𝐭(𝛑/𝒏)/2] = 𝐚²𝐂𝐨𝐭(𝛑/𝒏)/4. Determining the limit of the area of the polygon when 𝒏 goes to infinity: 𝐋𝐢𝐦 [𝐚²𝐂𝐨𝐭(𝛑/𝒏)/4] = 𝒏 → ∞ (𝐚²/4) 𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] = 𝒏 → ∞ 𝛑 is positive, so 𝛑/𝒏 when 𝒏 goes to infinity is also positive. When 𝒏 goes to infinity, 1/ 𝒏 goes to 0 by positive values. When 1/ 𝒏 goes to 0 by positive values, 𝐂𝐨𝐭(1/ 𝒏) goes to infinity. Therefore: 𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] = ∞ 𝒏 → ∞ 𝐚²/4 is always positive. Therefore: (𝐚²/4) 𝐋𝐢𝐦 [𝐂𝐨𝐭(𝛑/𝒏)] = ∞ 𝒏 → ∞ Conclusion: the "circle" is as big as you want it to be.

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@@SteveThePster The limiting shape is an apeirogon as someone pointed out in another comment.