On $\sum_{n=1}^\infty\frac{\mu\left(\lfloor \sqrt{n} \rfloor\right)-\mu\left(\lceil \sqrt{n} \rceil\right)}{n}$ and the Möbius function

Question

We denote the floor and ceiling functions, respectively, with $\lfloor x \rfloor$ and $\lceil x \rceil$.

We consider the Möbius function, and then the series $$\sum_{n=1}^\infty\frac{\mu\left(\lfloor \sqrt{n} \rfloor\right)-\mu\left(\lceil \sqrt{n} \rceil\right)}{n}.$$

When I've consider this series, the only simple calculation that I can to state is that those terms being a perfect square $m=k^2$ satisfy $$\lfloor \sqrt{m} \rfloor=\lceil \sqrt{m} \rceil=k,$$ and thus for these $m$'s their contribution in the series is $0=\frac{\mu\left(\lfloor \sqrt{m} \rfloor\right)-\mu\left(\lceil \sqrt{m} \rceil\right)}{m}$. But it is a simple observation, and doesn't tell me nothing about the convergence of the series.

I believe that it's an interesting example of series.

Question. Is it possible deduce that this series is convergent $$\sum_{n=1}^\infty\frac{\mu\left(\lfloor \sqrt{n} \rfloor\right)-\mu\left(\lceil \sqrt{n} \rceil\right)}{n}?$$ Many thanks.

Is not required an approximation, only is required the discussion about if the series is convergent or does diverge.

With this my code

sum (mu(floor(sqrt(n)))-mu(ceil(sqrt(n))))/n, from n=1 to 1000

Wolfram Alpha calculate an approximation and show us a graph.

Answer 1

$$\begin{eqnarray*}\sum_{m=1}^{M}\sum_{n=m^2}^{(m+1)^2-1}\frac{\mu\left\lfloor\sqrt{n}\right\rfloor-\mu\left\lceil \sqrt{n}\right\rceil}{n}&=&\sum_{m=1}^{M}\sum_{n=m^2+1}^{(m+1)^2-1}\frac{\mu\left\lfloor\sqrt{n}\right\rfloor-\mu\left\lceil \sqrt{n}\right\rceil}{n}\\&=&\sum_{m=1}^{M}\sum_{n=m^2+1}^{(m+1)^2-1}\frac{\mu(m)-\mu(m+1)}{n}\\&=&\sum_{m=1}^{M}\left(\mu(m)-\mu(m+1)\right)\left(H_{(m+1)^2-1}-H_{m^2}\right)\\(\text{summation by parts})\quad&=&-\mu(M+1)\left(H_{(M+1)^2-1}-H_{M^2}\right)\\&&+\sum_{m=1}^{M-1}\mu(m+1)\left(H_{(m+2)^2-1}-H_{(m+1)^2}-H_{(m+1)^2-1}+H_{m^2}\right)\\&=&O\left(\frac{1}{M}\right)+\sum_{m=2}^{M}\mu(m)\left(-\frac{2}{m^2}+\frac{2}{m^3}+O\left(\frac{1}{m^4}\right)\right)\end{eqnarray*}$$ is an absolutely convergent series, whose value is close to $-\frac{2}{\zeta(2)}+\frac{2}{\zeta(3)}$.

Answer 2

It seems that $\mu$ can be replaced by any bounded function $g$. Denote by $S_k$ the sum $$\sum_{i=1}^{2k}\frac 1{k^2+i}$$ It's easy to see that $S_k<2k/k^2=2/k$, hence $\lim_{k\to\infty}S_k=0$. Furthermore, $S_k-S_{k-1}<0$ for sufficiently large $k$ (this is a little fussy to show; instead one may just prove $|S_k-S_{k-1}|=O(1/k^2)$), hence $\sum_{n=1}^\infty|S_k-S_{k-1}|$ converges. Now note that \begin{align} \sum_{n=1}^{N^2}\frac{g(\lfloor n\rfloor)-g(\lceil n\rceil)}{n} & =\sum_{k=1}^{N-1}(g(k)-g(k+1))\sum_{i=1}^{2k}\frac 1{k^2+i}\\ & = \sum_{k=1}^{N-1}(g(k)-g(k+1))S_k\\ & = g(1)S_1+\sum_{k=2}^{N-1} g(k)(S_k-S_{k-1})-\lim_{N\to\infty}g(N)S_{N-1}\\&(\text{The last line is Abel summation formula}) \end{align} which is convergent since $g$ is bounded. For general $M$ with $N^2\le M<(N+1)^2$, just note that $$\left|\sum_{n=N^2+1}^M\frac{g(\lfloor n\rfloor)-g(\lceil n\rceil)}{n}\right|\le 2A\sum_{n=N^2+1}^M\frac 1 n\le \frac{2A(M-N^2)}{N^2}\le\frac{4A}{N}$$ tending to zero as $M\to\infty$.