Let $S_n = \sum_{k=1}^{n} \frac{n}{n^2 + kn + k^2}$ and $T_n = \sum_{k=0}^{n-1} \frac{n}{n^2 + kn +

Question

Let $S_n = \sum_{k=1}^{n} \frac{n}{n^2 + kn + k^2}$ and $T_n = \sum_{k=0}^{n-1} \frac{n}{n^2 + kn + k^2}$ for $n = 1, 2, 3, \dots$. Then,

Professor PC Thomas Classes · Answer

Both sums relate to the integral $I = \int_{0}^{1} \frac{1}{1 + x + x^2} dx = \int_{0}^{1} \frac{1}{(x + 1/2)^2 + (\sqrt{3}/2)^2} dx = \frac{2}{\sqrt{3}} \left[ an^{-1}\left(\frac{2x+1}{\sqrt{3}} ight) ight]_0^1 = \frac{2}{\sqrt{3}} \left( \frac{\pi}{3} - \frac{\pi}{6} ight) = \frac{\pi}{3\sqrt{3}}$. The function $f(x) = \frac{1}{1+x+x^2}$ is strictly decreasing on $[0, 1]$. $S_n$ is the right Riemann sum, which underestimates the area of a decreasing function ($S_n < I$). $T_n$ is the left Riemann sum, which overestimates the area ($T_n > I$). Thus, $S_n < \frac{\pi}{3\sqrt{3}}$ and $T_n > \frac{\pi}{3\sqrt{3}}$.