Zetas and positive even integers
Simplify to get:
2𝜋 4 5!
𝜁(4) = 𝜁(2) 2 −
𝜋 4 36
𝜋 4 60
𝜁(4)=
−
𝜋 4 90
𝜁(4)=
Now what about 𝜁(6)? By equating coefficients of 𝑥 7 get:
1 7!
1 𝜋 6
1 1 2
1 2 2
1 3 2
1 2 2
1 3
1 4 2
1 1 2
1 3 2
1 4 2
1 1 2
1 2 2
1 4 2
−
=−
(
×
×
+
×
×
+
×
×
+
×
×
+⋯) [𝟑]
Again, let the contents in the bracket be 𝑀 , and expand 𝜁(2) 3 :
3
1 1 2
1 2 2
1 3 2
𝜁(2) 3 =(
+
+
+⋯)
2
2
1 1 6
1 2 6
1 3 6
1 2 2
1 2 2
1 3 2
1 3 2
+⋯+3(1 2 ×
+1 2 ×
=
+
+
+1×(
)
+1×(
)
+⋯)
1 1 2
1 2 2
1 3 2
1 2 2
1 3
1 4 2
1 1 2
1 3 2
1 4 2
1 1 2
1 2 2
1 4 2
+6(
×
×
+
×
×
+
×
×
+
×
×
+⋯)
2
2
= 𝜁(6) + 6𝑀 + 3(1 2 × 1 2 2
1 2 2
1 3 2
1 3 2
+1 2 ×
+1×(
)
+1×(
)
+⋯) [𝟒]
Let the contents of the last bracket be 𝑁 , and expand 𝜁(2)𝜁(4) :
1 1 2
1 2 2
1 3 2
1 1 4
1 2 4
1 3 4
𝜁(2)𝜁(4) = (
+
+
+⋯)(
+
+
+⋯)
2
2
1 1 6
1 2 6
1 3 6
1 2 2
1 2 2
1 3 2
1 3 2
+⋯+(1 2 ×
+1 2 ×
=
+
+
+1×(
)
+1×(
)
+⋯)
= 𝜁(6) +𝑁 [𝟓]
Substitute [𝟓] into [𝟒] to get:
𝜁(2) 3 = 𝜁(6) + 6𝑀 + 3(𝜁(2)𝜁(4) − 𝜁(6))
Simplify to get:
1 6
1 2
1 3
𝜁(2) 3 −
𝑀 =
𝜁(2)𝜁(4) +
𝜁(6) [𝟔]
Substitute [𝟔] into [𝟑] to get:
1 7!
1 𝜋 6
1 6
1 2
1 3
𝜁(2) 3 −
−
=−
(
𝜁(2)𝜁(4) +
𝜁(6))
Simplify and solve to get 𝜁(6)= 𝜋 6 945
110
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