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, ...... Then

• A. S_n < \frac{\pi}{3\sqrt{3}}
• B. S_n > \frac{\pi}{3\sqrt{3}}
• C. T_n < \frac{\pi}{3\sqrt{3}}
• D. T_n > \frac{\pi}{3\sqrt{3}}

Answer 1

Answer: (A), (D)

A, D

Answer 2

The sums $S_n$ and $T_n$ can be rewritten as Riemann sums for the integral $\int_0^1 \frac{1}{1+x+x^2} dx$. $S_n = \sum_{k=1}^{n} \frac{1/n}{1 + (k/n) + (k/n)^2}$ is a right Riemann sum. $T_n = \sum_{k=0}^{n-1} \frac{1/n}{1 + (k/n) + (k/n)^2}$ is a left Riemann sum. The function $f(x) = \frac{1}{1+x+x^2}$ is strictly decreasing on $[0, 1]$ because its derivative $f'(x) = -\frac{1+2x}{(1+x+x^2)^2}$ is negative for $x \in [0, 1]$. The integral evaluates to $\int_0^1 \frac{1}{1+x+x^2} dx = \frac{\pi}{3\sqrt{3}}$. For a strictly decreasing function, the right Riemann sum is always less than the integral, and the left Riemann sum is always greater than the integral. Therefore, $S_n < \int_0^1 f(x) dx$ and $T_n > \int_0^1 f(x) dx$. This implies $S_n < \frac{\pi}{3\sqrt{3}}$ and $T_n > \frac{\pi}{3\sqrt{3}}$.