I'm looking for the asymptotic of the series $\displaystyle S(n)=\frac{1}{2}\sum_{k=1}^n\int_k^n\frac{dx}{x\sqrt{x(n-x)}}\,$ at $\,n\to\infty$
The first term can be found, for example, via switching to Riemann sums, changing the order of integration, and is equal to $\frac{\pi}{2}$. To find the next term I performed integration and got $$S(n)=\frac{1}{n}\sum_{k=1}^{n-1}\sqrt{\frac{n}{k}-1}=\frac{1}{\sqrt n}\sum_{k=1}^{n-1}\sqrt{\frac{1}{k}-\frac{1}{n}}=\frac{1}{\sqrt n}\sum_{k=1}^{n-1}f(k)\tag{1}$$ To get the flavour of the next asymptotic term I used the Euler-Maclaurin formula $$\sum_{k=1}^{n-1}f(k)\sim\int_1^{n-1}f(k)dk+\frac{1}{2}\big(f(1)+f(n-1)\big)+\frac{1}{12}\big(f'(n-1)-f'(1)\big)+ ...\tag{2}$$ At $n\to\infty$ the first term (the integral) gives $\frac{\pi}{2}\sqrt n-2$; other terms in (2) give non-zero values at $k=1$. Evaluating a couple of such terms, I got $$\sum_{k=1}^{n-1}f(k)\sim\frac{\pi}{2}\sqrt n-2+\frac{1}{2}+\frac{1}{24}=\frac{\pi}{2}\sqrt n-1.4583$$ The numeric evaluation at WolframAlpha for $n=1000$ gives $$ \sum_{k=1}^{n-1}f(k)-\frac{\pi}{2}\sqrt n\,\bigg|_{n=1000}=-1.46046$$ All this strongly resembles $\displaystyle \zeta\Big(\frac{1}{2}\Big)=-1.46035...\,\,$; at $\,n=10000\,$ we get even better agreement.
Questions:
- How can we prove that $\,\,\displaystyle \lim_{n\to\infty}\Big(\sum_{k=1}^n\sqrt{\frac{1}{k}-\frac{1}{n}}-\frac{\pi}{2}\sqrt n\Big)=\zeta\Big(\frac{1}{2}\Big)\,\,$?
- Can we get next asymptotic terms (at least, several of them) analytically ?