For positive integer $n$, define $f(n) = n + \frac{16+5n-3n^{2}}{4n+3n^{2}} + \frac{32+n-3n^{2}}{8n+3n^{2}} + \frac{48-3n-3n^{2}}{12n+3n^{2}} + ... + \frac{25n-7n^{2}}{7n^{2}}$. Then, the value of $\lim_{n \to \infty} f(n)$ is equal to

For positive integer n, define f(n) = n + (16+5n-3n²)/(4n+3n²) + (32+n-3n²)/(8n+3n²) + (48-3n-3n²)/(12n+3n²) + ... + (25n-7n²)/7n². Then, the value of lim_(n → ∞) f(n) is equal to $f(n) = n + \dfrac{16+5n-3n^2}{4n+3n^2} + \dfrac{32+n-3n^2}{8n+3n^2} + \ldots + \dfrac{25n-7n^2}{7n^2}$ $\lim_{n\to\infty} f(n) = ?$ $T_k = \dfrac{16k + (9-4k)n - 3n^2}{4kn + 3n^2},\qquad k = 1, 2, \ldots, n$ $k=n:\ \dfrac{16n + (9-4n)n - 3n^2}{4n^2 + 3n^2} = \dfrac{25n - 7n^2}{7n^2}\ \checkmark$ $T_k = -1 + \dfrac{16k+9n}{4kn+3n^2} = -1 + \dfrac{16k+9n}{n(3n+4k)}$ $f(n) = n + \sum_{k=1}^{n} T_k = n - n + \sum_{k=1}^{n} \dfrac{16k+9n}{n(3n+4k)}$ $= \sum_{k=1}^{n} \dfrac{16k+9n}{n(3n+4k)}$ $= \dfrac1n \sum_{k=1}^{n} \dfrac{16\frac{k}{n}+9}{3+4\frac{k}{n}} = \dfrac1n\sum_{k=1}^{n} g\!\left(\dfrac{k}{n}\right)$ $g(x) = \dfrac{16x+9}{3+4x}$ $\lim_{n\to\infty} f(n) = \displaystyle\int_0^1 \dfrac{16x+9}{3+4x}\,dx$ $\dfrac{16x+9}{3+4x} = 4 - \dfrac{3}{4x+3}$ $\int_0^1\!\left(4 - \dfrac{3}{4x+3}\right)dx = \left[4x - \dfrac34\ln(4x+3)\right]_0^1$ $= 4 - \dfrac34\ln\!\left(\dfrac73\right)$ $\boxed{4 - \dfrac34\log_e\!\left(\dfrac73\right)} \quad (2)$