Asymptotic estimate as $N \rightarrow \infty$ of $\sum\limits_{n = 1}^{N} \left\{{\frac{\left({n \pm 1}\right)}{{n}^{2}} N}\right\}$ 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.
 A: 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
$$
\sum\limits_{n = 1}^{N} \left\{{\frac{\left({n \pm 1}\right)}{{n}^{2}} N}\right\}
= (1 - \gamma) N + O(\sqrt N).
$$

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

*

*First part doesn't have too many summands.

*In the second sum we have $n > \frac{N}{\left[\sqrt N \right]} \geq \sqrt N$, which means we can "drop" braces: $\left\{ \frac{N}{n^2} \right\} = \frac{N}{n^2}$.

*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 $$
\left| \sum\limits_{ \frac{N}{\left[\sqrt N \right]} < n \leq N} [...] \right|
\leq \sum\limits_{k=1}^{\left[ \sqrt N \right] - 1} 2
= O(\sqrt N).
$$
