Fractional part distribution

Question

It is known that the distribution of $\{\sqrt{n} \}$, evaluated over the integer values of $n$, is uniform in the interval $[0,1)$. Let us consider the sum $$S(K)=\sum_{n=1}^K \left(\{\sqrt{n}\}-\frac{1}{2} \right) \sqrt{n}$$ I noted that the average value of $\sum_{K=1}^N S(K)$, calculated over all integers $N$ and for $N \rightarrow \infty$, converges to $1/4$. I suppose that this asymmetric distribution of $S(K)$ with respect to zero dipends on some properties of the distribution of $\sqrt{n} \mod 1$, but would be happy to better understand it by a formal proof.

Answer 1

Intuitively, we can consider the range of $n$ from $k^2$ to $(k+1)^2-1=k^2+2k$. Over this range $\lfloor \sqrt n \rfloor=k,$ so $\{\sqrt n\}=\sqrt n-k$. As $(k=\frac 12)^2=k^2+k+\frac 14$ the terms from $k^2$ through $k^2+k$ have $\{\sqrt n\}-\frac 12\lt 0$ and all the terms from $k^2+k+1$ through $k^2+2k$ have $\{\sqrt n\}-\frac 12\gt 0$. The positive ones get to multiply the larger $\sqrt n$'s. On the other hand, there is one less positive term.

Here is a failed attempt: We can write your sum with the limit of $K \to \infty$ as $$S=\sum_{n=1}^\infty \left(\{\sqrt{n}\}-\frac{1}{2} \right) \sqrt{n} =\sum_{m=1}^\infty\sum_{n=1}^{2m}\left(\sqrt{m^2+n}-m-\frac 12\right)\sqrt{m^2+n}$$ where we have broken the sum into pieces with each integer part of the square root. Designating each term in the outer sum as $S_m$ we have $$S_m=\sum_{n=0}^{2m}\left(\sqrt{m^2+n}-m-\frac 12\right)\sqrt{m^2+n}\\ =\sum_{n=0}^{2m}(m^2+n)-(m+\frac 12)\sqrt{m^2+n}\\ =2m^3+\frac 12(2m)(2m+1)-\sum_{n=0}^{2m}(m+\frac 12)\sqrt{m^2+n}$$ I tried turning the sum into an integral, but it is not accurate enough.