Mysterious subleading corrections to sums with internal dependence on limit Is there a standard method for finding expansions in $N$ of sums like
$$S(N)=\sum_{n=0}^N \sqrt{N^2-n^2}$$
beyond the first term?
It is easy to compute here that $$S(N)=N^2 \int_0 ^1 \sqrt{1-x^2} \mathrm
d x  + \mathcal O(N) = \pi N^2/4 + \mathcal O(N)$$
but finding $S(N) \approx \pi N^2 /4 + N/2$ is harder (I can do this by an unverified exchange of limits which fails at higher order), and at the moment I find the fact that the next correction seems to be a multiple of $\sqrt N$ pretty perplexing, and don't have an exact form for the coefficient. (Numerically I find $S \sim \pi N^2/4 + N/2 -0.2939955\ldots\sqrt N$, though this may not in fact be the correct form; I went up to $N=3,\!000,\!000$ plotting the ratio with $\sqrt N$).
Please note that I'm interested in generally applicable techniques for such a computation rather than this specific summation.
 A: Usually one can use the Euler-Maclaurin sum formula to get information like this.  Its first iteration takes the form
$$
\begin{align}
&\frac{1}{n} \sum_{k=0}^{n-1} f\left(\frac{k}{n}\right) \\
&\qquad = \int_0^1 f(x)\,dx + \frac{f(0)-f(1)}{2n} - \frac{1}{n} \int_0^1 \left(-nx - \lceil -nx\rceil - \frac{1}{2}\right)f'(x)\,dx,
\end{align}
$$
and when applied to your sum we get
$$
\frac{S(n)}{n^2} = \frac{\pi}{4} + \frac{1}{2n} + \frac{1}{n} \int_0^1 \left(-nx - \lceil -nx\rceil - \frac{1}{2}\right)\frac{x}{\sqrt{1-x^2}}\,dx.
$$
Now it just remains to estimate the integral.  To do this, we first expand the periodic factor in Fourier series as
$$
-nx - \lceil -nx\rceil - \frac{1}{2} = \frac{1}{\pi}  \sum_{k=1}^{\infty} \frac{\sin(2\pi k n x)}{k},
$$
so that
$$
\begin{align}
&\int_0^1 \left(-nx - \lceil -nx\rceil - \frac{1}{2}\right)\frac{x}{\sqrt{1-x^2}}\,dx \\
&\qquad = \frac{1}{\pi} \sum_{k=1}^{\infty} \frac{1}{k} \int_0^1 \sin(2\pi k n x) \frac{x}{\sqrt{1-x^2}}\,dx \\
&\qquad = \frac{1}{\pi} \sum_{k=1}^{\infty} \frac{1}{k} \left[\left. -\sin(2\pi k n x) \sqrt{1-x^2} \right|_0^1 + 2\pi k n \int_0^1 \cos(2\pi k n x) \sqrt{1-x^2}\,dx\right] \\
&\qquad = \frac{1}{\pi} \sum_{k=1}^{\infty} \frac{1}{k} \cdot 2\pi k n \int_0^1 \cos(2\pi k n x) \sqrt{1-x^2}\,dx \\
&\qquad = \frac{1}{2} \sum_{k=1}^{\infty} \frac{J_1(2\pi k n)}{k},
\end{align}
$$
where $J_1$ is the Bessel function of the first kind of order $1$.  This Bessel function has the asymptotic form
$$
J_1(x) = - \sqrt{\frac{2}{\pi x}} \cos\left(x+\frac{\pi}{4}\right) + O\left(x^{-3/2}\right)
$$
as $x \to \infty$, and since $\cos(2\pi k n + \pi/4) = 1/\sqrt{2}$ we get
$$
J_1(2\pi k n) = - \frac{1}{\pi \sqrt{2 k n}} + O\left((kn)^{-3/2}\right).
$$
It follows that
$$
\begin{align}
\sum_{k=1}^{\infty} \frac{J_1(2\pi k n)}{k} &= -\frac{1}{\pi \sqrt{2n}} \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} + O\left(n^{-3/2}\right) \\
& = -\frac{\zeta(3/2)}{\pi\sqrt{2n}} + O\left(n^{-3/2}\right)
\end{align}
$$
and hence that
$$
\int_0^1 \left(\lceil nx\rceil - nx - \frac{1}{2}\right)\frac{x}{\sqrt{1-x^2}}\,dx = -\frac{\zeta(3/2)}{\pi2^{3/2}\sqrt{n}} + O\left(n^{-3/2}\right).
$$
Therefore
$$
\frac{S(n)}{n^2} = \frac{\pi}{4} + \frac{1}{2n} - \frac{\zeta(3/2)}{\pi 2^{3/2} n^{3/2}} + O\left(n^{-5/2}\right).
$$
or

$$
S(n) = \frac{\pi}{4}n^2 + \frac{1}{2}n - \frac{\zeta(3/2)}{\pi 2^{3/2}} n^{1/2} + O\left(n^{-1/2}\right).
$$

Note that the constant you estimated on the $n^{1/2}$ term is
$$
- \frac{\zeta(3/2)}{\pi 2^{3/2}} \approx -0.293\ 995\ 518\ 793\ 519\ 291.
$$
