What about $\sum_{n=1}^\infty\frac{\mu(n)}{n}\psi\left(1+\frac{1}{n}\right)$ as $\frac{1}{2}$? For integers $n\geq 1$ let $\mu(n)$ the Möbius function, and let $\psi(z)$ the Digamma function, see its definition and how is denoted and typeset in Wolfram Language, if you need it from this MathWorld.
While I was doing calculations I found this problem:
Calculate a good approximation of $$\sum_{n=1}^\infty\frac{\mu(n)}{n}\psi\left(1+\frac{1}{n}\right).$$
My believe that why it is interesting is that Wolfram Alpha online calculator provide me approximations around $\frac{1}{2}$, with codes like this
sum mu(n)/n  PolyGamma[0, 1+1/n], from n=1 to 5000

Question. My belief is that defining $$S:=\sum_{n=1}^\infty\frac{\mu(n)}{n}\psi\left(1+\frac{1}{n}\right),$$ then $$S=\frac{1}{2}.$$ Can you provide us and justify a good approximation of $S$, or well do you know how discard that $S=\frac{1}{2}$? If do you know that it was in the literature, refers it. Many thanks.

 A: Since $\sum_{n=1}^\infty\frac{\mu(n)}{n}=0$, we have $$\sum_{n=1}^\infty\frac{\mu(n)}{n}\psi\left(1+\frac{x}{n}\right)=\sum_{n=1}^\infty\frac{\mu(n)}{n}\left[\psi\left(1+\frac{x}{n}\right)+\gamma\right],$$ and the series is uniformly convergent on $[0,1]$. Now
$$\sum_{n=1}^\infty\frac{\mu(n)}{n}\left[\psi\left(1+\frac{x}{n}\right)+\gamma\right]=-\sum_{n=1}^\infty\frac{\mu(n)}{n}\sum^\infty_{k=1}\zeta(k+1)\frac{(-x)^k}{n^k},$$ from the Taylor series of the Digamma function. For $|x|<1$, this is absolutely convergent, and we can change the order of summation, so we get $$-\sum^\infty_{k=1}\zeta(k+1)(-x)^k\sum_{n=1}^\infty\frac{\mu(n)}{n^{k+1}}=-\sum^\infty_{k=1}(-x)^k=\frac{x}{1+x},$$ since $$\sum_{n=1}^\infty\frac{\mu(n)}{n^{k+1}}=\frac1{\zeta(k+1)}.$$
The limit for $x\rightarrow1$ is $1/2,$ indeed.
A: This is not an answer but it is too long for a comment.
To confirm the result, I computed
$$S_k=\sum_{n=1}^{10^k}\frac{\mu(n)}{n}\psi\left(1+\frac{1}{n}\right)$$ and obtained the following decimal representations
$$\left(
\begin{array}{cc}
 k & S_k \\
 1 & 0.4606256866 \\
 2 & 0.4825538785 \\
 3 & 0.4974613056 \\
 4 & 0.5012018305 \\
 5 & 0.5002812270 \\
 6 & 0.4998842081
\end{array}
\right)$$ I gave up for $k=7$ (too long).
