Question

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

• A. $3+\frac{4}{3} \log _e 7$
• B. $4-\frac{3}{4} \log _e\left(\frac{7}{3}\right)$
• C. $4-\frac{4}{3} \log _e\left(\frac{7}{3}\right)$
• D. $3+\frac{3}{4} \log _{ e } 7$

Answer 1

Answer: (B)

$ f(n)=n+\displaystyle\sum_{r=1}^n \frac{16 r+(9-4 r) n-3 n^2}{4 r n+3 n^2}$
$f(n)=n+\displaystyle\sum_{r=1}^n \frac{(16 r+9 n)-\left(4 r n+3 n^2\right)}{4 r n+3 n^2}$
$ f(n)=n+\left(\displaystyle\sum_{r=1}^n \frac{16 r+9 n}{4 r n+3 n^2}\right)-n$
$\displaystyle\lim _{n \rightarrow \infty} f(n)=\displaystyle\lim _{n \rightarrow \infty} \sum \frac{16 r+9 n}{4 r n+3 n^2} $
$ =\displaystyle\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{\left(16\left(\frac{r}{n}\right)+9\right) \frac{1}{n}}{4\left(\frac{r}{n}\right)+3}$
$ =\int\limits_0^1 \frac{16 x+9}{4 x+3} d x=\int\limits_0^1 4 d x-\int\limits_0^1 \frac{3 d x}{4 x+3}$
$ =4-\frac{3}{4}(\ell n |4 x+3|)_0^1 $
$ =4-\frac{3}{4} \ell \operatorname{n} \frac{7}{3}$