\[\sum_{n=1}^{\infty} \mu (n)\]

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

Also ,

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

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

\[\sum_{n=1}^{\infty} \mu (n)\]

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

Also ,

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

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

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TopNewestTHe first sum diverges Link

The second sum follows suit. – Julian Poon · 1 year, 4 months ago

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Either ways, here's my approach:

Let \(f\) be a completely multiplicative function.

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

Through euler product.

Expanding the product gives

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

Putting it all together

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

Substituting \(f=1\) gives

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

\[\sum_{n>0}\mu(n)=0\]

Of course there is a lot of hand waving here. – Julian Poon · 1 year, 4 months ago

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– Aareyan Manzoor · 1 year, 4 months ago

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

– Julian Poon · 1 year, 4 months ago

Ok, so it divergesLog in to reply

@Julian Poon , @Aareyan Manzoor any modifications ? – Chinmay Sangawadekar · 1 year, 4 months ago

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– Chinmay Sangawadekar · 1 year, 4 months ago

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

– Julian Poon · 1 year, 4 months ago

THere is a high possibility my steps aren't justified, since I have assumed the sum converged.Log in to reply

@Julian Poon , @Aareyan Manzoor I'm waiting for your reply – Chinmay Sangawadekar · 1 year, 4 months ago

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Both sums diverges. – Aareyan Manzoor · 1 year, 4 months ago

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