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 $\{\}$ indicates the fractional part.

A similar question: Fractional part summation

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*}