# Fractional part summation and asymptotic expansion error

Let us consider the sum $$\displaystyle T_K=\sum_{n \geq \sqrt{K}}^{{K}} \left\{ \sqrt {n^2-K} \right\}$$ where $K$ is a positive integer and where $\{ \}$ indicates the fractional part. Its asymptotic expansion for $K \rightarrow \infty$ is

$$T_K=c K + O(\sqrt{K})$$

where

$$c= \frac{ 1+\gamma -\log(2) }{2}$$

and $\gamma$ is the Euler-Mascheroni constant. A nice demostration of this expansion is given here.

Now let us set $$A_K= c K- \frac{1}{2} \,\sqrt{K} -\frac{1}{8} \log(K)$$

where the last two terms of this quantity reflect the behaviour of the error term of the asymptotic expansion above. If we calculate the average value of the difference between $T_K$ and $A_K$ over all positive integers $K \leq N$, we have

$$\frac{1}{N} \sum_{K=1}^{N} (T_K- A_K) \approx 0.126....$$

for $N \rightarrow \infty \,\,$. I am interested in this constant term, for which there seems to be a rather slow convergence (the value above was calculated for $N=10^6\,$). In particular, I wonder whether this bias may have a closed-form expression. After some calculations, I found the quantity $$\gamma - \frac{1}{8} - \frac{\log(2)}{2}+\frac{1}{48}=0.1264754...\,$$ but I would like to have a formal confirmation of this.

• Try explaining in a few lines the asymptotic expansion.. – reuns Sep 24 '17 at 23:05
• Should it be $T_K=\frac{\gamma}{2} K + O(\sqrt{K})$ instead of $T_K=\frac{\gamma}{2} K + O(\sqrt{N})$? – Yuriy S Sep 26 '17 at 23:56
• What is $S_K$ here $\frac{1}{N} \sum_{K=1}^{N} (S_K- A_K) \approx 0.06....$? Should it be $T_K$ instead? – Yuriy S Sep 26 '17 at 23:59
• Typo corrected. Thanks. – Anatoly Sep 27 '17 at 13:30