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
 A: 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.
A: 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 $\frac{1}{5}$.
A: 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?".
