# A telescoping series with the Möbius function. Does a closed form exist?

With $\mu(n)$ the Möbius function, I experimented with the following function:

$$f(s)=\sum_{n=1}^\infty \frac{\mu(n)-\mu(n+1)}{n^s}$$

and found that:

$$f(1)= 1+\sum_{m=2}^\infty \left(1-\frac{1}{\zeta(m)}\right)$$

which is equal to Niven's constant.

Numerical evidence suggests that $f(s)$ is rapidly converging for $\Re(s) > 0$ and I wonder whether any other closed forms or expressions in series with $\zeta$ do exist for $s\ne 1$.

Thanks.

• I am interested in this kind of questions, my problem is that my mathematics are bads, thus I would like to encourage to you to explore more this kind of problems. Good luck. On the other hand @DanielFischer I don't know if were in the literature general theorems about series and the Möbius function, I've asked myself several times when $\sum_{n=1}^\infty a_n\mu(n)$ does converge (here $a_n$ is a infnite sequence)?
– user243301
Commented Jan 12, 2017 at 21:03

By summation by parts, $$f(s) = 1-\sum_{n\geq 1}\mu(n+1)\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right) \tag{1}$$ and working backwards: $$1-\frac{1}{\zeta(m)} = -\sum_{n\geq 2}\frac{\mu(n)}{n^m} = -\sum_{n\geq 1}\frac{\mu(n+1)}{(n+1)^m} \tag{2}$$ so: $$\sum_{m\geq 2}\left(1-\frac{1}{\zeta(m)}\right) = -\sum_{n\geq 1}\frac{\mu(n+1)}{n(n+1)}\tag{3}$$ re-proving your identity. If $s=2$, $$\frac{1}{n^2}-\frac{1}{(n+1)^2} = \frac{2n+1}{n^2(n+1)^2} = \sum_{m\geq 2}\frac{m}{(n+1)^{m+1}} \tag{4}$$ hence: $$f(2) = 1+\sum_{m\geq 2}m\left(1-\frac{1}{\zeta(m+1)}\right)\tag{5}$$ $$f(3) = 1+\sum_{m\geq 2}\frac{m(m+1)}{2}\left(1-\frac{1}{\zeta(m+2)}\right)\tag{6}$$ and so on.
• Generally with the terms $$(-1)^k\binom{-s}{k}\biggl(1 - \frac{1}{\zeta(k+s)}\biggr)$$ for $k \geqslant 1$. Commented Jan 11, 2017 at 20:53
• Thanks Jack & Daniel. Very helpful! Rewriting it gives: $$f(s)=1+\Gamma(1-s)\sum_{m=1}^{\infty}(-1)^m\left(\frac{1-\frac{1}{\zeta(m+s)}}{\Gamma(m+1)\,\Gamma(1-m-s)}\right)$$ which is valid for $\Re(s)>0$. Note that the RHS seems to provide an analytic continuation all the way to $\Re(s)=-3$ (were it hits the first pole). It also has real zeros. Not sure if complex zeros exist.