A rigurous proof to calculate $\lim_{r\to1}\sum_{n=0}^\infty\frac{\mu(n+1)}{(n+1)^3}r^n$, where $\mu(n)$ is the Möbius function

Question

Let $\mu(n)$ the Möbius function, you can see the definition in this MathWorld.

For $0<r<1$, I define $$f(r):=\sum_{n=0}^\infty\frac{\mu(n+1)}{(n+1)^3}r^n$$ and I've interested in calculate $$\lim_{r\to1}f(r),$$ (that is a notation for the limit $\lim_{r\to1^{-}}f(r)$).

I am interested about it after I've read Chapter 9 from Rudin, Functional analysis, International Series in Pure and Applied Mathematics, McGraw-Hill (1991), and create this example inspired in the introductory text of the chapter.

Question. Is well known that $$\sum_{n=1}^\infty\frac{\mu(n)}{n^3}=\frac{1}{\zeta(3)}$$ as a particular value of the Dirichlet series for the Möbius function. Is it possible to (justify and) calculate $$\lim_{r\to1}\sum_{n=0}^\infty\frac{\mu(n+1)}{(n+1)^3}r^n?$$

Thanks in advance.

Answer 1

For $0<r<1$ we have:

$$\Bigl|\frac{1}{\zeta(3)}-f(r)\Bigr|=\biggl|\sum\limits_{n=1}^\infty\frac{\mu(n)}{n^3}(1-r^{n-1})\biggr|\leq\biggl|\sum\limits_{n=2}^\infty\frac{\mu(n)}{n^3}\frac{1-r^{n-1}}{1-r}\biggr||1-r| \\ \leq (1-r)\sum\limits_{n=2}^\infty\frac{n-1}{n^3}=(1-r)\bigl(\zeta(2)-\zeta(3)\bigr)\to 0\quad\text{for}\quad r\to 1.$$