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I have to show $d|n

I do not really understand about this paragraph In this proof, what is meant by the leading term $p^n$ will dominate the remaining terms?

and also, how he conclude in the end that $\sigma(n)>0$? (Because I only see he proved that $n\sigma(n)<p^n$.)

Sil
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Nothing
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  • Keep your eye on the ball! The proof is about the remaining terms. That is, the terms with $d<n$. They show that $$\left\vert\sum_{d\mid n, d<n}\mu(n/d)p^d\right\vert<p^n.$$ Therefore $$|n\sigma(n)-p^n|=\left\vert\sum_{d\mid n, d<n}\mu(n/d)p^d\right\vert<p^n.$$ Therefore $n\sigma(n)>0$. – Jyrki Lahtonen Nov 21 '19 at 06:21
  • See the last part of this question. – Jyrki Lahtonen Nov 21 '19 at 06:27

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He is saying that $$n\sigma(n)=p^n+\sum_{d\mid n,d<n}\mu(n/d)p^d$$ and that $$\left|\sum_{d\mid n,d<n}\mu(n/d)p^d\right|\le\sum_{e=0}^{n-1}p^e =\frac{p^n-1}{p-1}<p^n.$$ This then implies that $n\sigma(n)>0$.

Angina Seng
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