# Problem with the Möbius Inversion Formula

(Edit) Can someone explain where the mistake is and what is wrong with the following "proof"?:

(i) $$\sum_{d\mid n}(\sigma_k(d))$$ = $$\sum_{d|n}(\tau(\frac{n}{d})d^k)$$ (True)

(ii) Define a function $$S(n)$$ such that $$\sum_{d\mid n}(\sigma_k(d)) = S(n) = \sum_{d|n}(\tau(\frac{n}{d})d^k)$$

(iii) The möbius inversion formula as it is stated in [Apostal, Analytic Number Theory, pg 32] says: If $$f(n) = \sum_{d|n}g(d)$$, then $$g(n) = \sum_{d|n}f(d)\mu(\frac{n}{d})$$

(iv)(a) $$S(n) = \sum_{d\mid n}(\sigma_k(d))$$, so we have $$\sigma_k(n) = \sum_{d|n}S(d)\mu(\frac{n}{d})$$

(iv)(b) $$S(n) = \sum_{d\mid n}(\tau(\frac{n}{d})d^k))$$, so we have $$\tau(\frac{n}{n})n^k = n^k = \sum_{d|n}S(d)\mu(\frac{n}{d})$$

(v) Therefore $$\sigma_k(n) = n^k$$

If anyone can tell me which line(s) is wrong and why it's wrong, that would be incredibly helpful. Thanks.

• What are $\sigma_k$ and $\tau$? – Aravind Aug 8 '20 at 17:22
• $\sigma_k(n)$ is the sum of the divisors of n, each raised to the kth power, and $\tau(n)$ is the number of divisors of n. – James Aug 8 '20 at 17:24
• How exactly did you use the MIF on the right side of your summation equation to get $n^k$? – Somos Aug 8 '20 at 17:27
• If you let both sums equal some function dependent on n, say S(n), by taking the right hand side term outside the summation you get $\tau(\frac{n}{n})*n^k = \sum_{d|n}(\mu(\frac{n}{d})*S(d))$ – James Aug 8 '20 at 17:29
• What do you mean with "taking the right hand side term outside the summation"? – Daniel Fischer Aug 8 '20 at 17:36

The mistake is in (iv)(b). In $$\sum_{d \mid n} \Biggl(\tau\biggl(\frac{n}{d}\biggr)d^k\Biggr) \tag{1}$$ the expression we sum over is not of the form $$f(d)$$, as it would need to be to apply the Möbius inversion formula (iii). The summands in $$(1)$$ depend on $$n$$ as well as on $$d$$, we have the form $$S(n) = \sum_{d \mid n} f(n,d)\,.$$
If we write things using the Dirichlet convolution using $$u \colon n \mapsto 1$$ and $$\nu_k \colon n \mapsto n^k$$ (thus we could write $$\nu_0$$ instead of $$u$$, but $$u$$ is a widely used name for that function), then the Möbius inversion formula can succinctly be stated as $$g = u \ast f \iff f = \mu \ast g\,. \tag{2}$$ The left hand side of (i) gives you $$S$$ in the form $$u \ast \sigma_k$$, precisely what $$(2)$$ is about, thus $$\sigma_k = \mu \ast S$$, or written in long form $$\sigma_k(n) = \sum_{d \mid k} \mu\biggl(\frac{n}{d}\biggr)S(d)\,,$$ by the Möbius inversion formula.
The right hand side of (i) however gives you $$S$$ in the form $$S = \tau \ast \nu_k$$, and $$(2)$$ says nothing about that form. A more general form of $$(2)$$ applies to that form however, and gives $$\nu_k = \tau^{-1} \ast S$$, or in long form $$n^k = \sum_{d \mid n} \tau^{-1}\biggl(\frac{n}{d}\biggr)S(d)\,,$$ where $$\tau^{-1}$$ is the Dirichlet inverse of $$\tau$$. (Since $$\tau = u\ast u$$ and $$\mu$$ is the Dirichlet inverse of $$u$$ we have $$\tau^{-1} = \mu \ast \mu$$.)
Let $$f_k(n)=n^k$$, then $$\sigma_k=1* f_k$$, thus $$1*\sigma_k=1*(1*f_k)=(1*1)*f_k=\tau*f_k$$. If you use Möbius inversion formula to $$\sigma_k=1*f_k$$, you get $$f_k=\mu*\sigma_k$$, not $$f_k=\sigma_k$$.
• It's the constant function $n\mapsto 1$. – Tuvasbien Aug 8 '20 at 17:51