We want to show \begin{align} \sum_{d|n,\ d>0}(\sigma(d)/d)\cdot \mu(n/d) =1/n , \end{align} where $\sigma(m)$ denotes the sum of all positive divisors of $m$ and where $\mu$ is the Möbius function.

The hint is that we can use the formula $\sigma(n)/n=\sum_{d|n,\ d>0} 1/d$.

I tried to pull out (1/d) from the sum and then convoluted the function to get \begin{align} \sum_{d|n,\ d>0}(\sigma(d)/d)\cdot \mu(n/d) =\sigma(n)/n\sum_{d|n,\ d>0} (\sigma(n/d))\cdot\mu(d) \end{align} and then tried to get $$\sum_{d|n,\ d>0} (\sigma(n/d))\cdot\mu(d)=1/\sigma(n)$$

But I can't figure out how to get there. Can anybody give me a hint?

  • 2
    $\begingroup$ Do you know the Möbius inversion formula? It's all you need. $\endgroup$ – FredH Mar 25 at 18:33
  • $\begingroup$ I know the formula but I don't get what to do with that $\endgroup$ – user657740 Mar 25 at 19:46
  • $\begingroup$ If $g(n) = \sum_{d\mid n} f(d)$, then $f(n) = \sum_{d\mid n} g(d)\mu(n/d)$. Let $f(n) = 1/n$. $\endgroup$ – FredH Mar 25 at 20:02

Hello and welcome to MSE.

Write $ \frac{σ(n)}{n} = (U*f)(n)$ , where $f(n)=\frac{1}{n}$, and $U(n)=1$ for all n.

You are then asked to find $((U*f) * (μ))$. Dirichlet convolution is associative and commutative, therefore, $(U*f) * (μ)= f*(U*μ)$, and $U* \mu$ is the identity of the Dirichlet convolution, therefore overall we get $f(n)$, as required.

(here * means dirichlet convolution (https://en.wikipedia.org/wiki/Dirichlet_convolution))

  • $\begingroup$ Why can you say $\sigma(n)/n=(1/n)*(1)?$ I'm not seeing the connection $\endgroup$ – user657740 Mar 25 at 19:44
  • $\begingroup$ That's because of the hint given to you, ie $\frac{\sigma(n)}{n}= \sum_{d|n}\frac{1}{d}$ which is equal to $ U * f$ using the convolution notation. $\endgroup$ – Locally unskillful Mar 25 at 20:07
  • $\begingroup$ @mupin By definition $\sigma(n) = \sum_{d | n} d = \sum_{d | n} n/d$ so $\sigma(n)/n = \sum_{d | n} 1/d= 1 \ast (1/n)$. Then $(\sigma(n)/n) \ast \mu =\sum_{l | n} \mu(n/l)\sum_{d | l} 1/d = \sum_{d | n} 1/d \sum_{l | n, d | l} \mu(n/l)$ with $k = n/l$ it is $= \sum_{d | n} \frac1d \sum_{k | n, k | n/d} \mu(k)= \sum_{d | n} \frac1d 1_{n/d= 1} = 1/n$ $\endgroup$ – reuns Mar 25 at 21:58

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