Limit of Definite Integral Sum Involving Logarithmic Terms

To evaluate the limit $\lim_{n \to \infty} f(n)$, we express the sum using its general term $T_k$ for $k = 1, 2, \dots, n$:

$T_k = \frac{16k + (9 - 4k)n - 3n^2}{4kn + 3n^2} = \frac{16k + 9n}{4kn + 3n^2} - 1$

Substituting $T_k$ into $f(n)$:

$f(n) = n + \sum_{k=1}^n T_k = n + \sum_{k=1}^n \left( \frac{16k + 9n}{4kn + 3n^2} - 1 \right) = \sum_{k=1}^n \frac{16k + 9n}{4kn + 3n^2} = \frac{1}{n} \sum_{k=1}^n \frac{16\left(\frac{k}{n}\right) + 9}{4\left(\frac{k}{n}\right) + 3}$

Expressing the limit as a definite integral:

$\lim_{n \to \infty} f(n) = \int_0^1 \frac{16x + 9}{4x + 3} \, dx = \int_0^1 \left( 4 - \frac{3}{4x + 3} \right) dx$

Evaluating the integral yields:

$\left[ 4x - \frac{3}{4} \log_e(4x + 3) \right]_0^1 = 4 - \frac{3}{4} \log_e\left(\frac{7}{3}\right)$

Hence, the correct option is (B).