Möbius Mania

n=1μ(n)\sum_{n=1}^{\infty} \mu (n)

1] Prove/Disprove , that the sum mentioned above converges.

Also ,

n=1(1)μ(n)\sum_{n=1}^{\infty}(-1) ^{\mu (n)}

2] Prove / Disprove , the sum mentioned above converges.

Note by A Former Brilliant Member
3 years, 7 months ago

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Both sums diverges.

Aareyan Manzoor - 3 years, 7 months ago

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THe first sum diverges Link

The second sum follows suit.

Julian Poon - 3 years, 7 months ago

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EDIT: My new thinking is that both sums does not converge or diverge since it's value fluctuates.

Either ways, here's my approach:

Let ff be a completely multiplicative function.

n>0f(n)=[p is prime(1f(p))]1\sum_{n>0}f(n)=\left[\prod_{p \text{ is prime}}(1-f(p))\right]^{-1}

Through euler product.

Expanding the product gives

p is prime(1f(p))=n>0μ(n)f(n)\prod_{p \text{ is prime}}(1-f(p))=\sum_{n>0}\mu(n)f(n)

Putting it all together

n>0f(n)=[n>0μ(n)f(n)]1\sum_{n>0}f(n)=\left[\sum_{n>0}\mu(n)f(n)\right]^{-1}

Substituting f=1f=1 gives

n>01=[n>0μ(n)]1\sum_{n>0}1=\left[\sum_{n>0}\mu(n)\right]^{-1}

n>0μ(n)=0\sum_{n>0}\mu(n)=0

Of course there is a lot of hand waving here.

Julian Poon - 3 years, 7 months ago

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Diverges means doesnt converge to a specific finite value, so it has to be eother one.

Aareyan Manzoor - 3 years, 7 months ago

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@Aareyan Manzoor Ok, so it diverges

Julian Poon - 3 years, 7 months ago

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@Julian Poon @Julian Poon , @Aareyan Manzoor any modifications ?

A Former Brilliant Member - 3 years, 7 months ago

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Till some point I also thought like this , but later , I left it as I thought it may be wrong.

A Former Brilliant Member - 3 years, 7 months ago

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@A Former Brilliant Member THere is a high possibility my steps aren't justified, since I have assumed the sum converged.

Julian Poon - 3 years, 7 months ago

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@Julian Poon @Julian Poon , @Aareyan Manzoor I'm waiting for your reply

A Former Brilliant Member - 3 years, 7 months ago

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