For positive integer n , define f n = n + 16 + 5 n - 3 n 2 4 n + 3 n 2 + 32 + n - 3 n 2 8 n + 3 n 2 + 48 - 3…

Given $f\left(n\right)=n+\frac{16+5n-3n^2}{4n+3n^2}+\frac{32+n-3n^2}{8n+3n^2}+\dots .+\frac{25n-7n^2}{7n^2}$

$=\left(\frac{16+5n-3n^2}{4n+3n^2}+1\right)+\left(\frac{32+n-3n^2}{8n+3n^2}+1\right)+\dots \dots +\left(\frac{25n-7n^2}{7n^2}+1\right)$

$f\left(n\right)=\frac{9n+16}{4n+3n^2}+\frac{9n+32}{8n+3n^2}+\dots \dots +\frac{25n}{7n^2}$

i.e. $f\left(n\right)=\sum _{r=1}^n\frac{9n+16r}{4rn+3n^2}=\frac{1}{n}\sum _{r=1}^n\frac{9+16\left(\frac{r}{n}\right)}{4\left(\frac{r}{n}\right)+3}$

Now, $\underset{n\to \infty }{\lim }f\left(n\right)={\int }_0^1\frac{9+16x}{4x+3}dx$

$={\int }_0^1\frac{\left(16x+12\right)-3}{4x+3}dx$

$={\left.\left(4x-\frac{3}{4}\ln \left|4x+3\right|\right)\right|}_0^1$ $=4-\frac{3}{4}\ln \frac{7}{3}$