+ B 3
E.g. B 1→
B 2
(Transfer one bulb to two)
=90 ゜ , OB 2
+ B 5
( ∠ B 5
⊥ B 4
B 2
→ B 4
OB 4
B 5
)
+ B 7
B 4→
B 6
We may discover that the arc length,
Length of arc OB 8
length is 3, length of arc OB 9
is 5.
Since the centers of circles with diameters 2 π , 4 π , 8 π
are on the same line. On the other hand, it is worth noting
that the line from o to central bulbs is always perpendicular to which the three bulbs lie. (e.g. OB 2 ⊥ B 4 B 5 ,
⊥ B 6
OB 4
B 7 ).This is because of each of these lines passes through the center of the circle we have drawn.
If we continue to double the radius of the circles. We can find that the arc from o to each bulb is getting
closer and closer to horizontal. Therefore, as we go to infinite, it is reasonable to assume that all the bulbs lie
on a horizontal line and the length of each line from a bulb to O equal to the correspond arc.
We can easily find out they form a line like this.
Hence: 1 2 π 2
=2 × （ 1 1 2
+ 1 3 2
+ 1
+ ⋯ ）
5 2
References.
π 2 8
= 1
+ 1 3 2
+ 1
+ ⋯
1 2
5 2
【 1 】 Gunter M Ziegler
A= 1 1 2
+ 1 2 2
+ 1 3 2
+ ⋯
【 2 】 Wikipedia
𝐴 4
= 1 2 2
+ 1
+ 1
+ ⋯
4 2
6 2
+ ⋯ = π 2 8
A − 𝐴 4
= 3 4
𝐴 = 1 1 2
+ 1 3 2
+ 1
5 2
Therefore A = π 2 6
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