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

Step 1: Recognise a Riemann sum.
A sum of the form $\frac{1}{n}\sum f\!\left(\frac{r}{n}\right)$ becomes $\int_0^1 f(x)\,dx$ as $n\to\infty$, with $\frac{r}{n}\to x$ and $\frac{1}{n}\to dx$.
Step 2: Write the integral.
Here $\frac{\pi r}{2n}=\frac{\pi}{2}\cdot\frac{r}{n}\to\frac{\pi}{2}x$, so \[ I=\int_0^1\sin^k\!\left(\tfrac{\pi}{2}x\right)\cos\!\left(\tfrac{\pi}{2}x\right)dx \]
Step 3: Substitute.
Let $u=\sin\!\left(\frac{\pi}{2}x\right)$, so $du=\frac{\pi}{2}\cos\!\left(\frac{\pi}{2}x\right)dx$, giving $\cos\!\left(\frac{\pi}{2}x\right)dx=\frac{2}{\pi}\,du$.
Step 4: Change the limits.
When $x=0$, $u=0$; when $x=1$, $u=\sin\frac{\pi}{2}=1$.
Step 5: Integrate.
\[ I=\frac{2}{\pi}\int_0^1 u^k\,du=\frac{2}{\pi}\left[\frac{u^{k+1}}{k+1}\right]_0^1 \]
Step 6: Evaluate.
\[ I=\frac{2}{\pi(k+1)} \] \[ \boxed{\dfrac{2}{\pi(k+1)}} \]