For positive integer \(n\), define $$f(n) = n + \frac{16+5n-3n^2}{4n+3n^2} + \frac{32+n-3n^2}{8n+3n^2}+\frac{48-3n-3n^2}{12n+3n^2}+\cdots + \frac{25n-7n^2}{7n^2}.$$ Then, evaluate \(\lim\limits_{n\to \infty}f(n)\).

Little Stories and Common Mistakes

I met with this problem from Aleams Barra in X. There are several answers and responses to this problem and common mistakes. Interestingly, this problem comes from JAA Advanced, an exam for students who want to take an undergraduate degree in India (if in Indonesia it is called UTBK).

We write \(\lim\limits_{n\to \infty}\) as \(\lim\) and \(f(n)\) as $$f(n) = n+\sum_{k=1}^n \frac{16k + (9-4k)n -3n^2}{4kn + 3n^2}.\tag{1}$$ To see where \( (1)\) come from, notice the pattern and use the arithmetic progression formula: $$16,32,48,\cdots; \quad 5,1,-3,\cdots;\quad 4,8,12,\cdots.$$ There are several common mistakes in dealing with this problem, using this property: $$\lim (a_n\pm b_n) = \lim a_n\pm \lim b_n.$$ This properties can be used if the limit value of \(\lim a_n\) and \(\lim b_n\) are exists. So, it's not true write $$\begin{align*}\lim f(n) &= \lim \left (n+\sum_{k=1}^n \frac{16k + (9-4k)n - 3n^2}{4kn+3n^2} \right )\\ &= \lim n + \sum_{k=1}^n \lim \left (\frac{16k + (9-4k)n -3n^2}{4kn+3n^2} \right ) \end{align*}$$ since \(\lim n\) doensn't exists. I thought the idea used the relationship Riemann Sums and definite integral: $$\int_a^b f(x)\;dx = \lim \sum_{k=1}^n f\left (a+\frac{b-a}{n}k\right )\frac{b-a}{n} .$$

Solution

Given \(f(n) = n + \sum\limits_{k=1}^n \frac{16k + (9-4k)n -3n^2}{4kn + 3n^2}\). Note that $$\begin{align*}f(n) &= n + \sum\limits_{k=1}^n \frac{16k + (9-4k)n -3n^2}{4kn + 3n^2} \\ &= \sum_{k=1}^n \left (1+\frac{16k +(9-4k)n -3n^2}{4kn+3n^2} \right )\\ &= \sum_{k=1}^n \frac{16k+9n}{4kn+3n^2} \\ &= \sum_{k=1}^n \frac{16\cdot\frac{k}{n} + 9}{4\cdot \frac{k}{n} + 3}\cdot \frac{1}{n}\end{align*}. $$ Take \(n\to\infty\) and applying relationship Riemann Sums with definite integral we have $$\lim f(n) = \int_0^1 \frac{16x+9}{4x+3}\;dx. $$ Let \(u=4x+3\), we have \(du=4\;dx\implies dx=\frac{du}{4}\). In other hand, \(x=\frac{u-3}{4}\) and we have $$16x+9 = 16\cdot \frac{u-3}{4}+9 = 4(u-3)+9 = 4u-3. $$ Thus $$\begin{align*}\int_0^1 \frac{16x+9}{4x+3} \;dx &= \int_3^{7} \frac{4u-3}{u}\;\frac{du}{4} \\ &= \frac{1}{4}\int_3^7 \left (4-\frac{3}{u}\right )\;du \\ &= \frac{1}{4}\left [4u - 3\ln(u) \right ]_3^7 \\ &= \frac{1}{4} \left [16 -3\ln(7) + 3\ln(3) \right ]\\ &= 4 +\frac{3}{4}\ln\left (\frac{3}{7}\right ).\end{align*} $$ Therefore, \(\lim f(n)=\boxed{4 +\frac{3}{4}\ln\left (\frac{3}{7}\right )}. \)