Asymptotic estimate as $N \rightarrow \infty$ of $\sum\limits_{n = 1}^{N} \left\{{\frac{\left({n \pm 1}\right)}{{n}^{2}} N}\right\}$

Question

Looking for the exact if possible, otherwise the asymptotic expansion and best estimate of the error terms as $N \rightarrow \infty$ of the two fractional sums $$\sum\limits_{n = 1}^{N} \left\{{\frac{\left({n \pm 1}\right)}{{n}^{2}} N}\right\}$$ My literature search has not found any examples similar to this. I have some material for the divisors terms such as $\left\lfloor{N/a}\right\rfloor^k$ and $\left\{{N/a}\right\}^k$. This is part of the calculation of the summations over the floor function of this argument.

From Benoit Cloitre. On the circle and divisor problem. November 2012 we have $$\lim_{x \rightarrow \infty} \sum_{n = 1}^{x} \left\lfloor{\frac{x}{{n}^{2}}}\right\rfloor \sim \zeta \left({2}\right) x + \zeta \left({\frac{1}{2}}\right) \sqrt{x} + O \left({{x}^{\theta}}\right)$$

Where $\theta = 1/4 + \epsilon$ is the estimated best error.

Answer 1

Asymptotic is $(1 - \gamma) N$, where $\gamma$ is Euler–Mascheroni constant.

Proof

For any $x, y$: $$ \begin{array}\\ \{x \pm y\} &= x \pm y - [x \pm y] \\ &= x \pm y - [[x] + \{x\} \pm [y] \pm \{y\}] \\ &= x - [x] \pm \{y\} - [\{x\} \pm \{y\}]. \end{array} $$

Now set $x = \frac Nn, y = \frac N{n^2}$ to break the sum apart:

$$ \sum\limits_{n = 1}^{N} \left\{{\frac{\left({n \pm 1}\right)}{{n}^{2}} N}\right\} = \underbrace{\sum\limits_{n = 1}^{N} \frac Nn}_{(1)} - \underbrace{\sum\limits_{n = 1}^{N} \left[ \frac Nn \right]}_{(2)} \pm \underbrace{\sum\limits_{n = 1}^{N} \left\{ \frac{N}{n^2} \right\} }_{(3)} - \underbrace{\sum\limits_{n = 1}^{N} \left[ \left\{ \frac Nn \right\} \pm \left\{\frac{N}{n^2} \right\} \right]}_{(4)}. $$

$(1)$ is Harmonic series, $(1) = N \ln N + \gamma N + \frac 12 + o(1)$.

$(2)$ is divisor summatory function, $(2) = N \ln N + N(2\gamma - 1) + O(\sqrt N)$.

$(3) = \underbrace{ \sum\limits_{n = 1}^{ \left[ \sqrt N \right]} \left\{ \frac{N}{n^2} \right\} }_{(3.1)} + \underbrace{ \sum\limits_{n = \left[ \sqrt N \right]+1}^{N} \left\{ \frac{N}{n^2} \right\} }_{(3.2)}. $

$ (3.1) \leq \sum\limits_{n = 1}^{ \left[ \sqrt N \right]} 1 \leq \sqrt N. $

$(3.2) = \sum\limits_{\left[ \sqrt N \right]+1}^{N} \frac{N}{n^2} \leq N \cdot \sum\limits_{\left[ \sqrt N \right]+1}^{N} \frac{1}{n (n-1)} = N \cdot \left( \sum\limits_{\left[ \sqrt N \right]+1}^{N} \frac{1}{n-1} - \frac{1}{n} \right) = N \left( \frac{1}{\left[ \sqrt N \right]} - \frac{1}{N} \right) \leq \frac {N}{\sqrt{N} + 1} - 1. $

$(4) = O(\sqrt N)$. Proof is very technical and is written below.

Putting $(1)$, $(2)$, $(3)$, $(4)$ together, and leaving only leading asymptotic terms we have

$$ \sum\limits_{n = 1}^{N} \left\{{\frac{\left({n \pm 1}\right)}{{n}^{2}} N}\right\} = (1 - \gamma) N + O(\sqrt N). $$

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Proving $(4) = O(\sqrt N)$

We want to show that $\sum\limits_{n = 1}^{N} \left[ \left\{ \frac Nn \right\} \pm \left\{\frac{N}{n^2} \right\} \right] = O(\sqrt N)$.

$$ \sum\limits_{1}^{N} [...] = \sum\limits_{1 \leq n \leq \frac{N}{\left[\sqrt N \right]} }[...] + \sum\limits_{ \frac{N}{\left[\sqrt N \right]} < n \leq N } [...], \\ $$

We split the sum in such a way, so that

1. First part doesn't have too many summands.
2. In the second sum we have $n > \frac{N}{\left[\sqrt N \right]} \geq \sqrt N$, which means we can "drop" braces: $\left\{ \frac{N}{n^2} \right\} = \frac{N}{n^2}$.
3. It will be convenient to work with the second sum later.

First sum is $O(\sqrt N)$ because the $[...]$ part equals either $-1$, $0$ or $1$: $$ \left| \sum\limits_{1 \leq n \leq \frac{N}{\left[\sqrt N \right]}} [...] \right| \leq \sum\limits_{1 \leq n \leq \frac{N}{\left[\sqrt N \right]}} 1 = O(\sqrt{N}) . $$

We'll split second sum even further, so that we can also "drop" braces for $\left\{ \frac Nn \right\}$:

$$ \sum\limits_{ \frac{N}{\left[\sqrt N \right]} < n \leq N} [...] = \sum\limits_{k=1}^{\left[ \sqrt N \right] - 1} \sum\limits_{\frac{N}{k + 1} < n \leq \frac Nk} [...]. $$

Note, that $\frac{N}{k + 1} < n \leq \frac Nk \implies k \leq \frac Nn < k + 1 \implies \left\{ \frac Nn \right\} = \frac Nn - k$.

$$ [...] = \left[ \left\{ \frac Nn \right\} \pm \left\{\frac{N}{n^2} \right\} \right] = \left[ \frac Nn - k \pm \frac {N}{n^2} \right] = \left[ N \frac{n \pm 1}{n^2} \right] - k. $$

When "$\pm$" is "$+$", the $[...]$ is either $0$ or $1$. We want to find for how many $n$ it is $1$.

$$ \left[ N \frac{n + 1}{n^2} \right] - k = 1 \iff N \frac{n + 1}{n^2} \geq k + 1 \iff \frac{k+1}{N}n^2 - n - 1 \leq 0, \\ \text{where} \; n \in \left( \frac{N}{k+1}; \frac Nk \right]. $$

Solving quadratic inequality gives $n \in \left( \frac{N}{k+1}; \frac{N}{k+1} \frac{1 + \sqrt{1 + 4 \frac{k+1}{N}}}{2} \right] $. Length of this semi-interval is

$$ \frac{N}{k+1} \frac{1 + \sqrt{1 + 4 \frac{k+1}{N}}}{2} - \frac{N}{k+1} = \frac{N}{k+1} \frac{-1 + \sqrt{1 + 4 \frac{k+1}{N}}}{2} = \frac{N}{k+1} \frac{-1 + 1 + 4 \frac{k+1}{N}}{2 \left(1 + \sqrt{1 + 4 \frac{k+1}{N}} \right) } = \frac{2}{1 + \sqrt{1 + 4 \frac{k+1}{N}}} < 1. $$

This means that at most $1$ integer $n$ can be inside that semi-interval.

When "$\pm$" is "$-$", the logic is similar, in that case there can be at most $2$ integer $n$ for which $[...] \neq 0$.

Finally, for the second sum we have $$ \left| \sum\limits_{ \frac{N}{\left[\sqrt N \right]} < n \leq N} [...] \right| \leq \sum\limits_{k=1}^{\left[ \sqrt N \right] - 1} 2 = O(\sqrt N). $$