New answers tagged mobius-function
5
votes
Accepted
On the identity $\sum_{n \leq x} \frac{1}{n} \sum_{k \leq \frac{x}{n}} \frac{\mu(k)}{k}=1$
Writing $m=nk$,
\begin{align*}
\sum_{n \le x} \frac{1}{n} \sum_{k \le x/n} \frac{\mu(k)}{k} &= \sum_{nk\le x} \frac1{nk} \mu(k) = \sum_{m\le x} \frac1m \sum_{k\mid m} \mu(k) = \sum_{m\le x} \...
- 85.1k
2
votes
Accepted
Asymptotic order of $\sum_{n>x}\frac{\mu (n)\ln n}{n^2}$
Your current bound $\sum_{n>x} n^{-3/2}$ would yield $O(x^{-1/2})$, which goes to $0$ slower than $O(\ln(x)/x)$. But you can get a better bound with a simple trick.
Observe that
$$\left|\sum_{n>...
- 27.7k
0
votes
Software for computing Möbius function of a poset
Sage and Macaulay2 are open source and provide functionality for computing the Möbius function of a partially ordered set.
Sage has the method ...
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