7
$\begingroup$

How can I prove that for every positive integer $n$ we have

\begin{equation*} \frac{n\pi}{4}−\frac{1}{\sqrt{8n}}<\frac{1}{2}+\sum_{k=1}^{n−1}\sqrt{1−\frac{k^2}{n^2}}? \end{equation*}

  • 1
    I would think to start by replacing the sum with the integral from 0 to n-1 and see if that works. No guarantees.2010-10-27
  • 0
    I have thought a proof using integrals as well as trying slicing a suitable part of circle but without success. Maybe one should try to do computations with details as I was unable to find a proof.2010-10-27
  • 1
    [Euler-Maclaurin summation][1] might help : We have : $\sum_{k=1}^{n} f(k) = \int_{1}^{n} f(x) dx + \frac{f(1)+f(n)}{2} + \int_{1}^{n} f'(x) (x - \lfloor x \rfloor - 1/2) dx $ for any decreasing $f$. Choosing $f = \sqrt{1-\frac{x^2}{n^2}}$ we get: $\sum_{k=1}^{n} \sqrt{1 -\frac{k^2}{n^2} } = \int_{1}^{n}\sqrt{1 - \frac{x}{n}^2} + \frac{1}{2} - \int_{1}^{n}( x - \lfloor x \rfloor - \frac{1}{2} ) \frac{x}{n \sqrt{n^2 - x^2}} $. [1]: http://calculus.nctu.edu.tw/~ocw/upload/fourier/supplement/euler%20summation.pdf2010-10-27

1 Answers 1

10

Write the inequality as $$\frac{\pi}{4} < \frac{1}{2n} + \frac{1}{n} \sum_{k=1}^{n-1} \sqrt{1-\left(\frac{k}{n}\right)^2} + \frac{1}{2n} \sqrt{\frac{1}{2n}}.$$ The left-hand side $\pi/4$ is the area of the part of the unit circle that lies in the first quadrant (below the curve $y=f(x)=\sqrt{1-x^2}$). We want to interpret the right-hand side as the area of a region $D$ which covers that quarter circle.

Note that $f$ is concave, so that its graph lies below any tangent line. Thus the trapezoid bounded by the lines $x=a-\epsilon$ and $x=a+\epsilon$ and by the $x$ axis and the tangent line through $(a,f(a))$ will cover the corresponding part of the circle: $$\int_{a-\epsilon}^{a+\epsilon} f(x) dx < 2\epsilon f(a).$$

Thus, taking $D$ to be the union of the following pieces does the trick:

  • A rectangle of height 1 between $x=0$ and $x=1/2n$.
  • Trapezoids as above, of width $\frac{1}{n}$ and centered at $x=k/n$ for $k=1,\ldots,n-1$.
  • A trapezoid as above, of width $\frac{1}{2n}$ and centered at $x=1-1/4n$. This last one has area $$\frac{1}{2n} f(1-1/4n) = \frac{1}{2n} \sqrt{\frac{1}{2n} - \frac{1}{16n^2}} < \frac{1}{2n} \sqrt{\frac{1}{2n}}.$$