Using mobius function compute sum (n k=1) of µ(k!) for each n in N?
Using mobius function for µ(k)
Compute sum (n k=1) of µ(k!) for each n in Natural Numbers
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Using mobius function for µ(k)
Compute sum (n k=1) of µ(k!) for each n in Natural Numbers
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Note that µ(n) = 0 if n is a square factor.
Then µ(4!) = µ(2^3 * 3) = 0.
==> µ(k!) = 0 for all k > 4, since 4 is a square factor of k!,
Hence, this sum equals
µ(1!) + µ(2!) + µ(3!) + 0
= µ(1) + µ(2) + µ(2 * 3)
= 1 + (-1) + (-1)^2
= 1.
I hope this helps!