Summation of square root fractional part

Question

Is there a way to calculate the following sum for large $c > K$, closed-form or approximation ?: $$ \sum_{n = 1}^{c}\left\{\sqrt{\,{K + n^{2}}\,}\right\} $$ where $K$ and $n$ are positive integers and $\left\{\cdots\right\}$ indicates the Fractional Part.

A similar question.

Answer 1

The given sum can be written as $$ \Sigma = \Sigma_1 - \Sigma_2 $$ where $$ \Sigma_1 = \sum_{n=1}^{c} \sqrt{n^2 + K} $$ and $$ \Sigma_2 = \sum_{n=1}^{c} \lfloor\sqrt{n^2 + K}\rfloor. $$ Now using $H_n = \log n + \gamma + \mathcal{O}(1/n)$ where $H_n$ is the $n$-th harmonic number, \begin{align*} \sum_{n=1}^{c} \sqrt{n^2 + K} - n &= \sum_{n=1}^c \frac{K}{\sqrt{n^2 + K} + n} \\ &\le \frac{K}{2} \sum_{n=1}^{c} \frac{1}{n} = \frac{K}{2}(\log c + \gamma + \mathcal{O}(1/c)). \end{align*} Hence $$ \Sigma_{1} = \sum_{n=1}^{c} \sqrt{n^2 + K} = \frac{1}{2}c(c+1) + \frac{K}{2}(\log c + \gamma) + \mathcal{O}(1/c) $$

Let $A = \lfloor\sqrt{K+1} - 1\rfloor$. Then, \begin{align*} \Sigma_{2} = \sum_{n=1}^{c} \lfloor\sqrt{n^2 + K}\rfloor &= \sum_{j=1}^{A}\sum_{\frac{K-(j+1)^2}{2(j+1)} < n \le \frac{K-j^2}{2j}}(n+j) + \sum_{\frac{K-1}{2} < n \le c}n \\ &= \sum_{n=1}^c n + \sum_{j=1}^{A}\sum_{\frac{K-(j+1)^2}{2(j+1)} < n \le \frac{K-j^2}{2j}}j\\ &= \frac{1}{2}c(c+1) + \sum_{j=1}^A j\left(\frac{3}{2} + \frac{K}{2j(j+1)}+ \mathcal{O}(1)\right)\\ &= \frac{1}{2}c(c+1) + \frac{3}{4}A(A+1) + \frac{K}{2}(H_A - 1) + \mathcal{O}\left(\sum_{j=1}^A j\right)\\ &= \frac{1}{2}c(c+1) + \frac{3}{4}A(A+1) + \frac{K}{2}(H_A - 1) + \mathcal{O}(1) \end{align*} since $\mathcal{O}\left(\sum_{j=1}^A j\right) = \mathcal{O}(K) = \mathcal{O}(1)$.

Thus,

\begin{align*} \Sigma &= \frac{K}{2}(\log c + \gamma) - \frac{3}{4}A(A+1) - \frac{K}{2}(H_A - 1) + \mathcal{O}(1).\\ \end{align*}

Answer 2

Partial solution: If we follow the example from Find a formula for $\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$. Then we have integer values for some integer $p$ where $p \le \sqrt{{k}^{2} + n} < p + 1 \Longleftrightarrow {p}^{2} \le {k}^{2} + n < \left({p + 1}\right)^{2} \Rightarrow p = \lfloor{\sqrt{{k}^{2} + n}}\rfloor$, or

$$ \sqrt{{p}^{2} - n} \le k < \sqrt{\left({p + 1}\right)^{2} - n} \Rightarrow 1 \le \lceil{\sqrt{{p}^{2} - n}}\rceil \le k \le \lceil{\sqrt{\left({p + 1}\right)^{2} - n}}\rceil - 1 \le c. $$

The maximum $p$ index is given by $\sqrt{{c}^{2} + n} < p + 1$ or ${p}_{max} = \lfloor{\sqrt{{c}^{2} + n}}\rfloor$ while the minimum $p$ index is ${p}_{min} = \lfloor{\sqrt{n +1}}\rfloor$. Also the lower $k$ index Iverson bracket constraint is needed to ensure that ${p}^{2} \ge n$, and the upper $k$ index constraint is to ensure that ${k}_{max} \le c$. So we can write

$$ \begin{aligned} {\Sigma}_{2} = \sum_{k = 1}^{c} \lfloor{\sqrt{{k}^{2} + n}}\rfloor {}={} & \sum_{p = \lfloor{\sqrt{n + 1}}\rfloor}^{\lfloor{\sqrt{{c}^{2} + n}}\rfloor} \sum_{k = \max \left({1, \lceil{\sqrt{{p}^{2} - n}}\rceil \left[{{p}^{2} \ge n}\right]}\right)}^{\min \left({c, \lceil{\sqrt{\left({p + 1}\right)^{2} - n}}\rceil - 1}\right)} \lfloor{\sqrt{{k}^{2} + n}}\rfloor \\ {}={} & \sum_{p = \lfloor{\sqrt{n + 1}}\rfloor}^{\lfloor{\sqrt{{c}^{2} + n}}\rfloor} p \sum_{k = \max \left({1, \lceil{\sqrt{{p}^{2} - n}\rceil} \left[{{p}^{2} \ge n}\right]}\right)}^{\min \left({c, \lceil{\sqrt{\left({p + 1}\right)^{2} - n}}\rceil - 1}\right)} 1 \\ {}={} & \sum_{p = \lfloor{\sqrt{n + 1}}\rfloor}^{\lfloor{\sqrt{{c}^{2} + n}}\rfloor} p \begin{bmatrix} 1 + \min \left({c, \lceil{\sqrt{\left({p + 1}\right)^{2} - n}}\rceil - 1}\right) \\ \mathop{-} \max \left({1, \lceil{\sqrt{{p}^{2} - n}}\rceil \left[{{p}^{2} \ge n}\right]}\right) \end{bmatrix}. \end{aligned} $$

At this point the remaining sums can be approximated using the Euler Maclaurin formula after dropping the floor and ceiling functions, and taking the dominant argument of the max and min argument lists. This will generate asymptotic expansions.

This solution holds for all values of $c$.

Answer 3

Upon further telescoping sum expansions and simplifications we can write for $c \ge \lfloor{(n-1)/2}\rfloor$ and with asymptotic expansions as $n \rightarrow \infty$

$${\Sigma}_{2} = \frac{1}{2}\, c \left({c + 1}\right) + \sum_{j = 1}^{A} \lfloor{\frac{K - {j}^{2}}{2\, j}}\rfloor \approx \frac{1}{2}\, {c}^{2} + \frac{1}{2}\, c + \frac{1}{4}\, K\, \log \left({K}\right) + \frac{1}{4} \left({2\, \gamma - 1}\right) K + \frac{11}{24} - \frac{49}{240}\, \frac{1}{K} + \frac{197}{1{,}260}\, \frac{1}{{K}^{2}} - \frac{421}{3{,}360}\, \frac{1}{{K}^{3}} + O \left({\frac{1}{{K}^{4}}}\right). $$

Which is exact. However, so far I can't find the correct summation for $c < \lfloor{(K-1)/2}\rfloor$.

See How to show that $\sum_{j = 1}^{A \left({n}\right)} \left\{{\frac{n - {j}^{2}}{2\, j}}\right\}$ satisfies Weyl's criterion. for a related question to solving the fractional part of the above floor sum. A change of variables, $K = n$.