# Compute $\lim_{n\to\infty}(n-(\arccos(1/n)+\cdots+\arccos(n/n)))$

How would you compute $$\lim_{n\to\infty}(n-(\arccos(1/n)+\cdots+\arccos(n/n)))?$$ If we choose $n=10000$ and compute it with W|A, we get $0.78833$ that is
suspiciously close to $\pi/4.$ I also discussed the limit in the chatroom. Thanks!

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The expression being subtracted is $n$ times the result of applying the trapezoidal rule to the integration of $\arccos x$ from $0$ to $1$, except for a contribution $\frac12(\arccos(0)-\arccos(1))=\pi/4$. The integral is $1$, and since the error of the trapezoidal rule is quadratic in the interval size, and thus falls off as $1/n^2$, the overall error goes to zero as $1/n$ with $n\to\infty$, so only the contribution $\pi/4$ remains.

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Interesting approach (+1). Thanks. – I'm an artist Nov 26 '12 at 11:40
@Chris'ssister: Your're welcome! – joriki Nov 26 '12 at 11:40
That the error part decreases as $1/N^2$ is decisive for the proof. Nice. :-) – Jayesh Badwaik Nov 26 '12 at 12:25
Actually, the error decays like $1/\sqrt{n}$. The problem arises near $x=1$ where the first derivative of $\arccos(x)$ blows up like $1/\sqrt{1-x}$. – robjohn Nov 26 '12 at 16:53

For $x\in[0,1]$, Taylor's Theorem says $$f\left(\frac{k-x}{n}\right) =f\left(\frac{k}{n}\right) -f'\left(\frac{k}{n}\right)\frac{x}{n} +O\left(\frac1{n^2}\right)$$ Summing in $k$ yields $$\sum_{k=1}^nf\left(\frac{k-x}{n}\right) =\sum_{k=1}^nf\left(\frac{k}{n}\right) -x\color{#C00000}{\sum_{k=1}^nf'\left(\frac{k}{n}\right)\frac1n} +O\left(\frac1n\right)$$ Integrating in $x$ over $[0,1]$ gives the first terms of the Euler-Maclaurin Sum Formula times $n$: $$\int_0^nf\left(\frac xn\right)\,\mathrm{d}x =n\int_0^1f(x)\,\mathrm{d}x =\sum_{k=1}^nf\left(\frac{k}{n}\right) -\frac12\color{#C00000}{(f(1)-f(0)) +O\left(\frac1{\sqrt{n}}\right)}$$ Setting $f(x)=1-\arccos(x)$ yields $$n-\sum_{k=1}^n\arccos\left(\frac kn\right)=\frac\pi4+O\left(\frac1{\sqrt{n}}\right)$$ Therefore, $$\lim_{n\to\infty}\left(n-\sum_{k=1}^n\arccos\left(\frac kn\right)\right)=\frac\pi4$$ Note about the red sum

When $f$ and $f'$ are monotonic, the error in the Riemann Sum approximation $$\sum_{k=1}^nf'\left(\frac{k}{n}\right)\frac1n=f(1)-f(0)$$ is less than $$\left|f(1)-f\left(1-\frac1n\right)\right|+\left|f\left(\frac1n\right)-f(0)\right|$$ In the case of $f(x)=1-\arccos(x)$, this error is $\sim\sqrt{\dfrac2n}$ .

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