Bounds on Riemann Sums for Definite Integral

To determine the relationship between $S_n$, $T_n$, and $\frac{\pi}{3\sqrt{3}}$, we rewrite the given sums in terms of a Riemann sum for a function $f(x)$.

The given expressions are:
$S_n = \sum_{k=1}^n \frac{n}{n^2 + kn + k^2} = \sum_{k=1}^n \frac{1}{n} \left(\frac{1}{1 + \frac{k}{n} + \left(\frac{k}{n}\right)^2}\right)$
$T_n = \sum_{k=0}^{n-1} \frac{n}{n^2 + kn + k^2} = \sum_{k=0}^{n-1} \frac{1}{n} \left(\frac{1}{1 + \frac{k}{n} + \left(\frac{k}{n}\right)^2}\right)$

Define the function:
$f(x) = \frac{1}{1 + x + x^2} \quad \text{for } x \in [0, 1]$

Differentiating $f(x)$ with respect to $x$:
$f'(x) = -\frac{2x + 1}{(1 + x + x^2)^2}$
Since $x \ge 0$, we have $2x + 1 > 0$, which implies $f'(x) < 0$ for all $x \in [0, 1]$.
Thus, $f(x)$ is a strictly decreasing function on the interval $[0, 1]$.

Divide the interval $[0, 1]$ into $n$ equal sub-intervals $\left[\frac{k-1}{n}, \frac{k}{n}\right]$ for $k = 1, 2, \dots, n$, each of width $\Delta x = \frac{1}{n}$.

Since $f(x)$ is strictly decreasing on each sub-interval $\left[\frac{k-1}{n}, \frac{k}{n}\right]$, we have:
$f\left(\frac{k}{n}\right) < f(x) < f\left(\frac{k-1}{n}\right) \quad \text{for all } x \in \left(\frac{k-1}{n}, \frac{k}{n}\right)$

Integrating over the interval $\left[\frac{k-1}{n}, \frac{k}{n}\right]$ gives:
$\frac{1}{n} f\left(\frac{k}{n}\right) < \int_{\frac{k-1}{n}}^{\frac{k}{n}} f(x)\, dx < \frac{1}{n} f\left(\frac{k-1}{n}\right)$

Summing these inequalities from $k = 1$ to $n$:
$\sum_{k=1}^n \frac{1}{n} f\left(\frac{k}{n}\right) < \sum_{k=1}^n \int_{\frac{k-1}{n}}^{\frac{k}{n}} f(x)\, dx < \sum_{k=1}^n \frac{1}{n} f\left(\frac{k-1}{n}\right)$

This simplifies to:
$S_n < \int_0^1 f(x)\, dx < T_n$

Now, we evaluate the definite integral $\int_0^1 f(x)\, dx$:
$\int_0^1 \frac{1}{x^2 + x + 1}\, dx = \int_0^1 \frac{1}{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}\, dx$
$= \left[ \frac{2}{\sqrt{3}} \arctan\left(\frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) \right]_0^1$
$= \frac{2}{\sqrt{3}} \left[ \arctan\left(\frac{2x + 1}{\sqrt{3}}\right) \right]_0^1$
$= \frac{2}{\sqrt{3}} \left( \arctan(\sqrt{3}) - \arctan\left(\frac{1}{\sqrt{3}}\right) \right)$
$= \frac{2}{\sqrt{3}} \left( \frac{\pi}{3} - \frac{\pi}{6} \right) = \frac{2}{\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi}{3\sqrt{3}}$

Substituting the value of the integral back into the inequality, we obtain:
$S_n < \frac{\pi}{3\sqrt{3}} < T_n$

Thus:
$S_n < \frac{\pi}{3\sqrt{3}} \quad \text{and} \quad T_n > \frac{\pi}{3\sqrt{3}}$

Therefore, the correct options are A and D.