# Möbius Mania

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

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. Note by A Former Brilliant Member
3 years, 11 months ago

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Both sums diverges.

- 3 years, 11 months ago

The second sum follows suit.

- 3 years, 11 months ago

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 $f$ 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.

- 3 years, 11 months ago

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

- 3 years, 11 months ago

Ok, so it diverges

- 3 years, 11 months ago

@Julian Poon , @Aareyan Manzoor any modifications ?

- 3 years, 11 months ago

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

- 3 years, 11 months ago

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

- 3 years, 11 months ago