Asymptotic behavior of weighted sums involving the fractional part function

Question

Currently, I am studying the asymptotic behavior of sums of the form \begin{equation}\label{eq1}\tag{1} \sum_{k=1}^{n-1} f(n-k) \left\{ \frac{n}{k} \right\} \end{equation}

In this context, based on limited computations (up to $5-6$ digits), it seems that the following limit might equal $\frac{\zeta(2)}{2}$: \begin{equation}\label{eq2}\tag{2} \lim_{n \to \infty} \left( \zeta(2) n +\log(n) - \frac{1}{n} \sum_{k=1}^{n-1} \left(1 - \frac{k}{n}\right)^{-3} \left\{ \frac{n}{k} \right\} \right) \end{equation}

My approach so far has been to try to evaluate this limit using the following formula, valid for any positive integer $j>0$: \begin{equation}\label{eq3}\tag{3} I(j):=\int_{0}^{1} t^{j} \left\{ \frac{1}{t} \right\} \, dt = \frac{1}{j} - \frac{\zeta(j+1)}{j+1} \end{equation} where $I(0)=1-\gamma$ (with $\gamma$ being the Euler-Mascheroni constant).

This seems natural when considering the development: \begin{equation}\label{eq4}\tag{4} \frac{1}{(1-x)^3} = \sum_{i \geq 0} \frac{(i+1)(i+2)}{2}x^i \end{equation}

By splitting the sum over $i$ and interchanging sums for $i \leq m$, I've established: \begin{equation}\label{eq5}\tag{5} \sum_{i=0}^m \frac{(i+1)(i+2)}{2}I(j) = \frac{m}{2} + \log(m) - \frac{1}{2}\zeta(2) + o(1) \end{equation}

However, I'm unable to bridge the gap between these intermediate results and the original limit.

I would appreciate:

1. Any insights on how to proceed from here or elsewhere!
2. References related to the asymptotic behavior of sums of type \eqref{eq1}
3. Suggestions for alternative approaches to prove the conjectured limit

PS : This question is part of a more general conjecture for integer values $m \geq 3$. Namely, I conjecture that the following limit exists: $$ C(m) = \lim_{n \rightarrow \infty} \left( \log(n) + \sum_{k=2}^{m-1} \zeta(k) n^{k-1} - \frac{1}{n} \sum_{k=1}^{n-1} \left(1 - \frac{k}{n}\right)^{-m} \left\{ \frac{n}{k} \right\} \right) $$ I suspect that $C(3) = \frac{\zeta(2)}{2}$, but I currently have no idea for $m \geq 4$. Understanding the case $C(3)$ could provide valuable insights toward addressing this broader conjecture.

Moreover, if this conjecture holds, I would like to know if there exist constants $\theta_m$ such that $$ \sum_{k=1}^{n-1} \left(1 - \frac{k}{n}\right)^{-m} \left\{ \frac{n}{k} \right\} = \sum_{k=2}^{m-1} \zeta(k) n^{k} + n \log(n) - C(m) n + O\left(n^{\theta_{m}}\right). $$ Through numerical experiments with $ m=3$ and assuming $C(3) = \frac{\zeta(2)}{2}$, it seems that $\theta_3 =1/2$ works and even $\frac{1}{4} + \epsilon$ might be suitable. This is reminiscent of the Dirichlet divisor problem, which seeks the optimal $\theta$ such that $$ \sum_{k=1}^{n-1} \left\{ \frac{n}{k} \right\} = \left(1 - \gamma\right) n + O\left(n^{\theta}\right). $$

My guess is that $\theta_m=1/4+\epsilon$ is also the optimal value for all $m\geq3$.

PS 2 We can formulate a broader conjecture by considering the following four sums defined for integers $ m \geq 0 $:

$$ S_{m}(n) = \sum_{k=1}^{n-1} \left( 1 - \frac{k}{n} \right)^{-m} \left\{ \frac{n}{k} \right\} $$

$$ T_{m}(n) = \sum_{k=1}^{n-1} (-1)^{k-1} \left( 1 - \frac{k}{n} \right)^{-m} \left\{ \frac{n}{k} \right\} $$

$$ U_{m}(n) = \sum_{k=1}^{n-1} \left( 1 - \frac{k}{n} \right)^{-m} \left\Vert \frac{n}{k} \right\Vert $$

$$ V_{m}(n) = \sum_{k=1}^{n-1} (-1)^{k-1} \left( 1 - \frac{k}{n} \right)^{-m} \left\Vert \frac{n}{k} \right\Vert $$

where $ \{x\} $ denotes the fractional part of $ x $, and $ \|x\| $ denotes the distance to the nearest integer.

---

Conjecture

If $\theta $ denotes the optimal exponent in the error term of the Dirichlet divisor problem, which corresponds to:

$$ S_{0}(n) = \left( 1 - \gamma \right) n + O\left(n^{\theta}\right), $$

then the following results hold:

For $ S_{m}(n) $:

• $ S_{1}(n) = n + O\left(n^{\theta}\right) $,
• $ S_{2}(n) = n \log n + n + O\left(n^{\theta}\right) $,
• For $ m \geq 3 $: $$ S_{m}(n) = \sum_{k=2}^{m-1} \zeta(k) n^k + n \log n - C_{m} n + O\left(n^{\theta}\right), $$ where thanks to H. Cohen we have the conjectured value: $$ C_{m} = \frac{1}{m-1} \sum_{k=2}^{m-1} \binom{m-1}{k} \zeta(k) + \sum_{k=2}^{m-2} \frac{1}{k}. $$

---

For $ T_{m}(n) $:

• $T_{0}(n) = O\left(n^{\theta}\right) $,
• $T_{1}(n) = O\left(n^{\theta}\right) $,
• For $ m \geq 2 $: $$ T_{m}(n) = \sum_{k=1}^{m-1} \eta(k) n^k + O\left(n^{\theta}\right), $$ where $\eta(k)=\sum_{j=1}^\infty \frac{(-1)^{j-1}}{j^k}$.

---

For $U_{m}(n) $:

• $U_{0}(n) = \log\left(\frac{4}{\pi}\right)n + O\left(n^{\theta}\right) $,
• $ U_{1}(n) = \log(2) n + O\left(n^{\theta}\right) $,
• $ U_{2}(n) = n \log n - K_{2} n + O\left(n^{\theta}\right) $, with $ K_{2} = -0.4324\ldots $,
• For $m \geq 3 $: $$ U_{m}(n) = \sum_{k=2}^{m-1} \zeta(k) n^k + n \log n - K_{m} n + O\left(n^{\theta}\right), $$ where $K_{3} = 1.9538\ldots , K_{4} = 4.9367\ldots $, etc.

---

For $V_{m}(n)$:

• $ V_{0}(n) = O\left(n^{\theta}\right) $,
• $V_{1}(n) = O\left(n^{\theta}\right) $,
• For $m \geq 2 $: $$ V_{m}(n) = \sum_{k=1}^{m-1} \eta(k) n^k + O\left(n^{\theta}\right). $$

Remark: $T_m(n)$ and $V_m(n)$ behave similarly. $S_m$ and $ U_m $ are almost similar, differing only by the constants $C_m$ and $K_m$.

Answer 1

Not an answer but a conjectural answer for the value of $C(m)$ supported by extensive numerical evidence:

$$C(m)=\dfrac{1}{m-1}\sum_{2\le k\le m-1}\binom{m-1}{k}\zeta(k)+\sum_{2\le k\le m-2}\dfrac{1}{k}$$

For instance $C(3)=\zeta(2)/2$, $C(4)=\zeta(3)/3+\zeta(2)+1/2$, $C(5)=\zeta(4)/4+\zeta(3)+3\zeta(2)/2+5/6$.