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1 solution
Notice that 2 2 = 4 ∣ n ! for n ≥ 4 . Hence μ ( n ! ) = 0 if n > 4 . It follows that,
n = 1 ∑ ∞ μ ( n ! ) = μ ( 1 ! ) + μ ( 2 ! ) + μ ( 3 ! ) = 1 .