# What about the convergence of $\sum_{n=1}^\infty\frac{(\mu(n))^n}{n}$, where $\mu(n)$ is the Möbius function?

Let for integers $n\geq 1$ the Möbius function $\mu(n)$. See this Wikipedia to know the definition of this arithmetic function MathWorld.

I don't know if this question was in the literature, refers the paper if it is well-known and I try find it.

Question. What about the convergence of $$\sum_{n=1}^\infty\frac{\mu(n)^n}{n}?$$ Many thanks.

See if you want this code using Wolfram Alpha online calculator

sum mu(n)^n/n, from n=1 to 10000

• Many thanks who upvoted the question. – user243301 Sep 23 '17 at 18:36
• Do you realize not simplifying $\mu(n)^n$ is more than ridiculous ? – reuns Sep 23 '17 at 19:03
• Yes, many thanks. The argument was more complicated because one need to combine with the densities of those numbers that are/aren't squarefree. Also many thanks for your contribution below @reuns – user243301 Sep 23 '17 at 20:41

This is yet another answer. Here we show that the sum diverges to $\infty$. We begin by noting that

$$\mu(n)^n = \begin{cases} \mu(n), & \text{if n is odd}, \\ 1, & \text{if n is even and squarefree}. \end{cases}$$

Now let $\mathbb{P}_{\mathrm{odd}}$ be the set of odd primes. Then for any $\Re(s) > 1$ we have

\begin{align*} \sum_{n=1}^{\infty} \frac{\mu(n)^n}{n^s} &= \sum_{\text{$n$ odd}} \frac{\mu(n)}{n^s} + \sum_{\text{$n$ even, squarefree}} \frac{1}{n^s} \\ &= \prod_{p \in \mathbb{P}_{\mathrm{odd}}} \left( 1 - \frac{1}{p^s}\right) + \frac{1}{2^s} \prod_{p \in \mathbb{P}_{\mathrm{odd}}} \left( 1 + \frac{1}{p^s}\right). \end{align*}

Although this computation is not justified at $s = 1$ at this moment, we also know that

$$\sum_{n=1}^{N} \frac{\mu(n)^n}{n} = \sum_{\substack{n \leq N \\ \text{n odd}}} \frac{\mu(n)}{n} + \sum_{\substack{n \leq N \\ \text{n even, squarefree}}} \frac{1}{n}.$$

Since we know that the latter sum tends to

$$\sum_{\text{n even, squarefree}} \frac{1}{n} = \frac{1}{2} \prod_{p \in \mathbb{P}_{\mathrm{odd}}} \left( 1 + \frac{1}{p}\right) = \infty,$$

it suffices to show the following slightly stronger claim:

Claim. $\displaystyle \sum_{\text{$n$odd}} \frac{\mu(n)}{n}$ converges to $0$.

Basics of Lambert series. To show the claim above, we breifly make a digress. Notice that if $a_n = \mathcal{O}(n^{\sigma})$, then for any $t > 0$ we have

$$\sum_{n=1}^{\infty} a_n \frac{nt}{e^{nt} - 1} = t \sum_{n=1}^{\infty} n a_n \left( \sum_{k=1}^{\infty} e^{-nkt} \right) = t \sum_{n=1}^{\infty} \left( \sum_{d\mid n} d a_d \right) e^{-nt}. \tag{1}$$

As usual, we have the following Abelian theorem:

Theorem. If $\sum_{n=1}^{\infty} a_n$ converges to $S$, then $\text{(1)}$ also converges to $S$ as $t \to 0^+$.

Moreover, we have the following Lambert Tauberian theorem.

Theorem. If either $|n a_n| \leq C$ or $n a_n \geq -C$ and $\text{(1)}$ converges to $S$ as $t \to 0^+$, then $\sum_{n=1}^{\infty} a_n$ also converges to $S$.

Now we apply above theorems to our case. First, notice that for any $s \in \mathbb{C}$ we have

$$t \sum_{\text{n odd}} \frac{\mu(n)}{n^s} \frac{nt}{e^{nt} - 1} = t \sum_{n=1}^{\infty} \Bigg( \prod_{\substack{p \in \mathbb{P}_{\mathrm{odd}} \\ p \mid n}} \left( 1 - \frac{1}{p^{s-1}} \right) \Bigg) e^{-nt}. \tag{2}$$

Then by the Abelian theorem and $\text{(1)}$, for $\Re(s) > 1$ we have

$$\frac{1}{(1 - 2^{-s})\zeta(s)} = \prod_{p \in \mathbb{P}_{\mathrm{odd}}} \left( 1 - \frac{1}{p^s}\right) = \sum_{\text{n odd}} \frac{\mu(n)}{n^s} = \lim_{t\to 0^+} t \sum_{\text{n odd}} \frac{\mu(n)}{n^s} \frac{nt}{e^{nt} - 1}$$

Taking $s \to 1^+$, the above quantity vanishes. On the other hand, exploiting the non-negativity and monotonicity of the RHS of $\text{(2)}$ for $s \geq 1$, we find that

$$0 \leq t \sum_{\text{n odd}} \frac{\mu(n)}{n} \frac{nt}{e^{nt} - 1} \leq t \sum_{\text{n odd}} \frac{\mu(n)}{n^s} \frac{nt}{e^{nt} - 1}.$$

Thus taking $t \to 0^+$ followed by $s \to 1^+$,

$$\lim_{t \to 0^+} t \sum_{\text{n odd}} \frac{\mu(n)}{n} \frac{nt}{e^{nt} - 1} = 0.$$

By the Lambert Tauberian theorem, this implies the claim as desired.

• Many thanks you are very generous. Now all uses can to read your answer. – user243301 Sep 23 '17 at 21:32

This is $$\sum_{n\text{ odd}}\frac{\mu(n)}n+\sum_{n\text{ even, squarefree}}\frac{1}n=S_1+S_2$$ say. Well maybe it is, if both $S_1$ and $S_2$ are convergent sums. But $$S_2=\prod_{p\text{ odd, prime}}\left(1+\frac1p\right)$$ which diverges. If $S_1$ converges, then your sum diverges (to $\infty$).

I think $S_1$ does. It is known (corollary of Prime Number Theorem) that $$\sum_{n=1}^\infty\frac{\mu(n)}{n}$$ converges (to zero). Separating off the even and odd terms gives $$\sum_{n=1}^\infty\frac{\mu(n)}{n} =\sum_{n\text{ odd}}\frac{\mu(n)}n+\sum_{n\text{ even}}\frac{\mu(n)}n =\sum_{n\text{ odd}}\frac{\mu(n)}n-\frac12\sum_{n\text{ odd}}\frac{\mu(n)}n$$ but I'm not sure of this: maybe $\sum_{n\text{ odd}}\mu(n)/n$ diverges? If it converges, it must converge to zero, so I'm no longer sure of anything. Anyway, the question reduces to "is series $S_1$ convergent?".

• I doubted if it was a good question, many thanks you for your answer. Tomorrow I am going to study your reasoning. – user243301 Sep 23 '17 at 18:26
• $\sum_{n=1}^\infty \mu(2n-1) (2n-1)^{-s} = \frac{1}{1-2^{-s}} \frac{1}{\zeta(s)} = \sum_{k=0}^\infty \sum_{n=1}^\infty \mu(n)(2^k n)^{-s}$ thus $\sum_{n=1}^\infty \frac{\mu(2n-1)}{ 2n-1}= \lim_{N \to \infty} \sum_{k \ge 0} 2^{-k} \sum_{n \ge N/2^k} \frac{\mu(n)}{n} = 0$ – reuns Sep 23 '17 at 18:46
• @reuns, It took a while to understand your argument, but I finally got it. (+1) Also, a bit nitpicking but you have a typo: you should have $n \leq N/2^k$ instead. $$\sum_{\substack{1\leq n \leq N \\ \text{n odd}}} \frac{\mu(n)}{n} = \sum_{k \geq 0} \frac{1}{2^k} \sum_{1 \leq n \leq N/2^k} \frac{\mu(n)}{n}.$$ – Sangchul Lee Sep 23 '17 at 21:08

I would expect a divergent behaviour. $\mu(n)$ is non-zero only if $n$ is squarefree. With such assumption, $\mu(n)^n=1$ if $n$ is even or has an even number of prime factors, $\mu(n)^n = -1$ iff $n$ is odd and with an odd number of prime factors, which is less likely than the previous event. Since the partial sums of $\sum_{n\geq 1}\mu(n)^n$ are expected to be unbounded, by Kronecker's lemma the given series is expected to be divergent, behaving like $$\text{Conjecture.}\qquad \sum_{n=1}^{N}\frac{\mu(n)^n}{n}\approx C \log(N).$$ Explanation: $\frac{6}{\pi^2}$ is the density of squarefree-numbers, $\sum_{n=1}^{N}\frac{1}{n}\approx \log(N)$, if $n$ is a squarefree number then $\mu(n)^n$ is expected to be $1$ with probability $\frac{3}{4}$ and $-1$ with probability $\frac{1}{4}$. Numerically, $C$ seems to be around $\frac{1}{5}$.

• I doubted if it was a good question, many thank you for your answer. Tomorrow I study it. – user243301 Sep 23 '17 at 18:26
• @reuns: it should be assuming the studied constraints are perfectly independent, but a numerical simulation suggests they are not. – Jack D'Aurizio Sep 23 '17 at 18:45
• I meant $\sum |\mu(2n-1)| (2n-1)^{-1-s} = \frac{\zeta(s+1)}{\zeta(2(s+1))} \frac{1}{1+2^{-s-1}}$ so $\sum_{2n-1 \le x} \frac{|\mu(2n-1)|}{2n-1} \sim \text{Res}(\frac{\zeta(s+1)}{\zeta(2(s+1))} \frac{1}{1+2^{-s-1}}\frac{x^s}{s},0) = C \log x+B+o(1)$ – reuns Sep 23 '17 at 19:00