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Two proofs:
1 . − ζ ( k ) > 1 , ( ∀ k ≥ 2 , k ∈ N ) ⇒ k = 2 ∑ ∞ ζ ( k ) = + ∞
2 . −
If k = 2 ∑ ∞ ζ ( k ) converged then k → ∞ lim ζ ( k ) = 0 , but k → ∞ lim ζ ( k ) = 1 , so the previous sum doesn't converge and therefore, due to 1 it diverges to + ∞