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*}
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*}
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: