# Value of Möbius series involving $\log(n+1)$

Let $$\mu$$ be the Möbius function. It is known that

$$\sum_{n \geq 1} \frac{\mu(n)}{n} \log n = -1$$

I'm looking for a closed-form expression of the similar series

$$\sum_{n \geq 1} \frac{\mu(n)}{n} \log (n+1) \approx -0.63\dots$$

I tried to manipulate the series in various ways by using the Mercator series and asymptotic series of the logarithm and other known Möbius series (e.g. in terms of the Riemann zeta function and its derivatives), but couldn't find a closed-form.

$$\begin{array}\\ f(m) &=\sum_{n \geq 1} \dfrac{\mu(n)}{n} \log (n+m)\\ &=\sum_{n \geq 1} \dfrac{\mu(n)}{n} \log (n+m)-\sum_{n \geq 1} \dfrac{\mu(n)}{n} \log (n)+\sum_{n \geq 1} \dfrac{\mu(n)}{n} \log (n)\\ &=\sum_{n \geq 1} \dfrac{\mu(n)}{n} (\log (n+m)-\log (n)) -1\\ &=\sum_{n \geq 1} \dfrac{\mu(n)}{n} \log (1+m/n) -1\\ &=\sum_{1 \le n \le m} \dfrac{\mu(n)}{n} \log (1+m/n)+\sum_{n \geq m+11} \dfrac{\mu(n)}{n} \log (1+m/n) -1\\ &=\sum_{1 \le n \le m} \dfrac{\mu(n)}{n} \log (1+m/n)+\sum_{n \geq m+1} \dfrac{\mu(n)}{n} \sum_{k=1}^{\infty} (-1)^{k+1}\dfrac{m^k}{n^k} -1\\ &=\sum_{1 \le n \le m} \dfrac{\mu(n)}{n} \log (1+m/n)+\sum_{k=1}^{\infty} (-1)^{k+1}m^k\sum_{n \geq m+1} \dfrac{\mu(n)}{n} \dfrac{1}{n^k} -1\\ &=\sum_{1 \le n \le m} \dfrac{\mu(n)}{n} \log (1+m/n)+\sum_{k=1}^{\infty} (-1)^{k+1}m^k\sum_{n \geq m+1} \dfrac{\mu(n)}{n^{k+1}} -1\\ \text{so}\\ f(1) &=\sum_{n \geq 1} \dfrac{\mu(n)}{n} \log (n+1)\\ &=\sum_{1 \le n \le 1} \dfrac{\mu(n)}{n} \log (1+1/n)+\sum_{k=1}^{\infty} (-1)^{k+1}\sum_{n \geq 2} \dfrac{\mu(n)}{n^{k+1}} -1\\ &=\log (2)+\sum_{k=1}^{\infty} (-1)^{k+1}(-1+\sum_{n \geq 1} \dfrac{\mu(n)}{n^{k+1}}) -1\\ &=\log (2)+\sum_{k=1}^{\infty} (-1)^{k+1}(-1+\dfrac1{\zeta(k+1)}) -1\\ &=\log (2)-1+\sum_{k=1}^{\infty} (-1)^{k+1}\dfrac{1-\zeta(k+1)}{\zeta(k+1)} \\ &=\log (2)-1+\sum_{k=1}^{\infty} (-1)^{k}\dfrac{\zeta(k+1)-1}{\zeta(k+1)} \\ \end{array}$$
Note: Ths sum converges since $$\zeta(k)-1 \approx \dfrac1{2^k}$$ so $$\dfrac{\zeta(k+1)-1}{\zeta(k+1)} \approx \dfrac1{2^{k+1}}$$.