For positive integer n, define

$f\left(n\right)=n+\frac{16+5n-3n^2}{4n+3n^2}+\frac{32+n-3n^2}{8n+3n^2}+\frac{48-3n-3n^2}{12n+3n^2}+\cdots +\frac{25n-7n^2}{7n^2}$

Then, the value of $\underset{n\to \infty }{\lim }f\left(n\right)$ is equal to

The correct answer is $4-\frac{3}{4}{\log }_e\left(\frac{7}{3}\right)$.

Step 1: Identify the General Term

Let us look at the terms added to $n$ in $f\left(n\right)$:

Term 1: $\frac{16+5n-3n^2}{4n+3n^2}$ (k = 1)
Term 2: $\frac{32+n-3n^2}{8n+3n^2}$ (k = 2)
Term 3: $\frac{48-3n-3n^2}{12n+3n^2}$ (k = 3)

We observe the following pattern for the k-th term:

— Numerator constant: $16k$

— Coefficient of n: $9-4k$ (starts at 5, decreases by 4 each step)

— Coefficient of n²: always $-3$

— Denominator: $4kn+3n^2$

So the k-th fraction is:

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

We can verify the last term at k = n:
Numerator: $16n+(9-4n)n-3n^2=16n+9n-4n^2-3n^2=25n-7n^2$ ✓
Denominator: $4n·n+3n^2=7n^2$ ✓

So the sum runs from k = 1 to k = n, and:

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

Step 2: Simplify Each Fraction

Notice that the numerator of each fraction can be rewritten as:

$16k+(9-4k)n-3n^2=-(4kn+3n^2)+(9n+16k)$

So each fraction becomes:

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

Step 3: Substitute Back into f(n)

$f\left(n\right)=n+\sum _{k=1}^n\left[-1+\frac{9n+16k}{n(3n+4k)}\right]$

$=n-n+\sum _{k=1}^n\frac{9n+16k}{n(3n+4k)}$

$f\left(n\right)=\frac{1}{n}\sum _{k=1}^n\frac{9+16\frac{k}{n}}{3+4\frac{k}{n}}$

Step 4: Recognise as a Riemann Sum and Convert to an Integral

This is a Riemann sum with $t=\frac{k}{n}$ and step size $\frac{1}{n}$, over the interval [0, 1]. Therefore:

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

Step 5: Evaluate the Integral

Use the substitution $u=3+4t$, so $du=4dt$ and $t=\frac{u-3}{4}$.

The limits change: when $t=0$, $u=3$; when $t=1$, $u=7$.

The numerator transforms:

$9+16t=9+16·\frac{u-3}{4}=9+4(u-3)=4u-3$

So the integral becomes:

${\int }_3^7\frac{4u-3}{u}·\frac{d}{u}=\frac{1}{4}{\int }_3^7\left(4-\frac{3}{u}\right)du$

Integrate term by term:

$=\frac{1}{4}\left[4u-3{\log }_e\left|u\right|\right]{}_3^7$

$=\frac{1}{4}\left[\right(4·7-3{\log }_{e}7)-(4·3-3{\log }_{e}3\left)\right]$

$=\frac{1}{4}[28-3{\log }_{e}7-12+3{\log }_{e}3]$

$=\frac{1}{4}[16+3{\log }_e\frac{3}{7}]$

$=4+\frac{3}{4}{\log }_e\frac{3}{7}$

Since ${\log }_e\frac{3}{7}=-{\log }_e\frac{7}{3}$, we get:

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

Therefore:

$\underset{n\to \infty }{\lim }f\left(n\right)=4-\frac{3}{4}{\log }_e\left(\frac{7}{3}\right)$