Zetas and positive even integers
𝑥 2 𝜋 2
𝑥 2 2 2 𝜋 2
𝑥 2 3 2 𝜋 2
sin(𝑥) 𝑥
lim 𝑥→0
= lim 𝑥→0
𝐴(1−
)(1−
)(1−
)…
Left-hand side approaches 1 using small angle approximation. On the right, each term except 𝐴 approaches 1 as 𝑥 goes to 0 . So 𝐴 must be 1 .
Therefore, we have:
𝑥 2 𝜋 2
𝑥 2 2 2 𝜋 2
𝑥 2 3 2 𝜋 2
sin(𝑥) = 𝑥 (1 −
)(1−
)(1−
)…
And
𝑥 3 3!
𝑥 5 5!
𝑥 7 7!
𝑥 2 𝜋 2
𝑥 2 2 2 𝜋 2
𝑥 2 3 2 𝜋 2
𝑥−
+
−
+⋯= 𝑥(1−
)(1−
)(1−
)…
Now, we can obtain the value of 𝜁(2) by equating the coefficients of 𝑥 3 .
1 3!
1 𝜋 2
1 2 2
1 3 2
−
=−
(1+
+
+⋯)
Multiplying both sides by −𝜋 2 to get:
𝜋 2 6
1 2 2
1 3 2
=(1+
+
+⋯)= 𝜁(2)
Unfortunately, it gets much harder from here. To obtain the value for 𝜁(4) , we need to look at the coefficients of 𝑥 5 .
1 5!
1 𝜋 4
1 2 2
1 3 2
1 2 2
1 3 2
=
(1×
+1×
+
×
+⋯)[𝟏]
At first glance, the right-hand side looks like a pile of mess that has nothing to do with 𝜁(4) . Let’s call the sum of everything in the bracket on the right-hand side 𝑀 for mess.
However, consider the expansion of 𝜁(2) 2 , we have:
2
1 1 2
1 2 2
1 3 2
𝜁(2) 2 =(
+
+
+⋯)
2
2
2
1 1 2
1 2 2
1 3 2
1 2 2
1 3 2
1 2 2
1 3 2
=(
)
+(
)
+(
)
+⋯+2(1×
+1×
+
×
+⋯)
= 𝜁(4) + 2𝑀
So, we have
1 2 (𝜁(2) 2 − 𝜁(4)) [𝟐]
𝑀 =
Substitute into [𝟐] into [𝟏] to get:
1 5!
1 2𝜋 4
(𝜁(2) 2 −𝜁(4))
=
109
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