An application of Mobius Inversion $\sum_{d \mid n} \mu(\frac{n}{d})\nu(d) = 1$

Question

Show that $\sum_{d \mid n} \mu(\frac{n}{d})\nu(d) = 1$, for any positive integer n. Where $\mu$ denotes the Mobius function defined by $\mu(n)=(-1)^{s}$ if $n=p_{1} \dotsc p_{s}$ for distinct primes $p_{1} \dotsc p_{s}$ and $\mu(n)=0$ otherwise, and $\nu(n)$ denotes the number of divisors of $n$.

I think I have to apply the Mobius Inversion Theorem somehow or use properties of the Dirichlet product, but I'm not sure. Any help would be greatly appreciated.

Can you prove it's true for primes? prime powers? Can you prove it's a multiplicative function? — Gerry Myerson, Oct 09 '16 at 10:06
OK I have managed to prove all three things and after a bit of work the result should follow? — jackwo, Oct 09 '16 at 11:53
Well, that's the point of multiplicative functions – once you can evaluate them at prime powers, you can evaluate them everywhere. — Gerry Myerson, Oct 09 '16 at 11:59
One formula for $\nu$ is $\nu(n) = \sum_{d|n} 1$. What does Mobius inversion do to that? — B. Goddard, Oct 09 '16 at 12:56
This is simply $$\frac{1}{\zeta(s)}\times \zeta(s)^2 = \zeta(s).$$ — Marko Riedel, Oct 09 '16 at 20:21

score 2 · Answer 1 · edited Jan 15 '17 at 18:22

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$\nu(n)= \sum_{d|n} 1$ => According to the Moebius Inversion Formula $\sum_{d|n} μ(n/d)\nu(d) = 1$

edited Jan 15 '17 at 18:22

Martin Sleziak

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answered Jan 15 '17 at 18:09

Chain Markov

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An application of Mobius Inversion $\sum_{d \mid n} \mu(\frac{n}{d})\nu(d) = 1$

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