1 − 5 + 3 − 5 + 5 − 5 + 7 − 5 + . . . 1 − 5 + 2 − 5 + 3 − 5 + 4 − 5 + 5 − 5 + 6 − 5 + 7 − 5 + . . . =
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Let the sum be S . Then S = 1 + 3 2 S (breaking even and odd terms of the numerator and taking 2 − 5 common from the even terms) , or S = 3 1 3 2
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Q = 1 − 5 + 3 − 5 + 5 − 5 + 7 − 5 + ⋯ 1 − 5 + 2 − 5 + 3 − 5 + 4 − 5 + 5 − 5 + 6 − 5 + 7 − 5 + ⋯ = 1 − 5 + 2 − 5 + 3 − 5 + ⋯ − ( 2 − 5 + 4 − 5 + 6 − 5 + ⋯ ) ζ ( 5 ) = 1 − 5 + 2 − 5 + 3 − 5 + ⋯ − 2 − 5 ( 1 − 5 + 2 − 5 + 3 − 5 + ⋯ ) ζ ( 5 ) = ζ ( 5 ) − 2 − 5 ζ ( 5 ) ζ ( 5 ) = 1 − 3 2 1 1 = 3 1 3 2 Riemann zeta function ζ ( n ) = k = 1 ∑ ∞ k n 1
Reference: Riemann zeta function