Question

Evaluate \[ \lim_{n\to\infty}\frac{1}{n} \left[ \tan^2\frac{\pi}{4n} +\tan^2\frac{2\pi}{4n} +\cdots+ \tan^2\frac{n\pi}{4n} \right]. \]

• A. \(1\)
• B. \(0\)
• C. \(-1\)
• D. \(\dfrac{4-\pi}{\pi}\)

Hint:

Whenever a limit contains \[ \frac1n\sum_{r=1}^{n}f\!\left(\frac{r}{n}\right), \] recognize it as a Riemann sum and convert it directly into \[ \int_0^1 f(x)\,dx. \]

Answer 1

The Correct Option is D

Concept: Convert the given limit into a Riemann sum. If \[ \lim_{n\to\infty} \frac1n \sum_{r=1}^{n} f\!\left(\frac{r}{n}\right), \] then \[ = \int_0^1 f(x)\,dx. \]

Step 1: Rewrite the given sum. \[ L = \lim_{n\to\infty} \frac1n \sum_{r=1}^{n} \tan^2\left(\frac{r\pi}{4n}\right). \] Let \[ f(x)=\tan^2\left(\frac{\pi x}{4}\right). \] Then \[ L = \int_0^1 \tan^2\left(\frac{\pi x}{4}\right)\,dx. \]

Step 2: Evaluate the integral. Put \[ t=\frac{\pi x}{4}. \] Then \[ dx=\frac{4}{\pi}\,dt. \] When \[ x=0,\quad t=0, \] and when \[ x=1,\quad t=\frac{\pi}{4}. \] Therefore, \[ L = \frac{4}{\pi} \int_0^{\pi/4} \tan^2 t\,dt. \] Using \[ \tan^2 t=\sec^2 t-1, \] \[ L = \frac{4}{\pi} \int_0^{\pi/4} (\sec^2 t-1)\,dt. \] \[ = \frac{4}{\pi} \Big[\tan t-t\Big]_0^{\pi/4}. \] \[ = \frac{4}{\pi} \left( 1-\frac{\pi}{4} \right). \] \[ = \frac{4-\pi}{\pi}. \] Therefore, \[ \boxed{ \lim_{n\to\infty}\frac{1}{n} \left[ \tan^2\frac{\pi}{4n} +\tan^2\frac{2\pi}{4n} +\cdots+ \tan^2\frac{n\pi}{4n} \right] = \frac{4-\pi}{\pi} } \] \[ \boxed{\text{Answer = (D)}} \]