Question

\( \lim_{n \rightarrow \infty} \frac{1}{n}\sum_{r=1}^{n}\sin^{k}\left(\frac{\pi r}{2n}\right)\cos\left(\frac{\pi r}{2n}\right) = \)

• A. \( \frac{1}{k+1} \)
• B. \( \frac{\pi}{2(k+1)} \)
• C. \( \frac{2}{\pi(k+1)} \)
• D. \( \frac{2}{k+1} \)

Hint:

Always remember the scaling coefficient multiplier. Differentiating the nested trigonometric term \( \sin\left(\frac{\pi}{2}x\right) \) brings out a factor of \( \frac{\pi}{2} \), which flips to become \( \frac{2}{\pi} \) when placed outside the integral.

Answer 1

The Correct Option is C

Concept: A Riemann sum limit can be elegantly converted into a definite integral using the standard mapping transformations: \[ \frac{r}{n} \rightarrow x, \quad \frac{1}{n} \rightarrow dx, \quad \lim_{n \rightarrow \infty}\sum \rightarrow \int_{0}^{1} \]

Step 1: Converting the limit string into definite integral form.
Applying the transformations to our given expression: \[ I = \int_{0}^{1} \sin^{k}\left(\frac{\pi}{2}x\right)\cos\left(\frac{\pi}{2}x\right) \, dx \]

Step 2: Integrating using substitution.
Let \( u = \sin\left(\frac{\pi}{2}x\right) \implies du = \frac{\pi}{2}\cos\left(\frac{\pi}{2}x\right) \, dx \implies \cos\left(\frac{\pi}{2}x\right) \, dx = \frac{2}{\pi} \, du \). Let us find the new integration boundaries:

• Lower limit: When \( x = 0 \implies u = \sin(0) = 0 \)

• Upper limit: When \( x = 1 \implies u = \sin\left(\frac{\pi}{2}\right) = 1 \)
Substituting these into the integral equation: \[ I = \int_{0}^{1} u^k \left(\frac{2}{\pi}\right) \, du = \frac{2}{\pi} \left[ \frac{u^{k+1}}{k+1} \right]_{0}^{1} = \frac{2}{\pi(k+1)} \] This matches option (C) perfectly.