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)

$4-\frac{3}{4} \log _e\left(\frac{7}{3}\right)$

Answer 2

The correct option is (B): $4-\frac{3}{4} \log _e\left(\frac{7}{3}\right)$