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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestTHe first sum diverges Link

The second sum follows suit.

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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 $f$ 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.

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

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

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

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@Julian Poon , @Aareyan Manzoor any modifications ?

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Both sums diverges.

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