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.

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.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.