Question

Evaluate the limit as n → ∞ of the sum
1/n [sin (π/n) +sin (2π)/n +.....+ sin (2nπ)/n]

Answer 1

Answer on Question #69112 – Math – Real Analysis

Question

Evaluate the limit as $n \to \infty$ of the sum

$$1/n \left[ \sin \frac{(\pi}{n}) + \sin \frac{(2\pi)}{n} + \cdots + \sin \frac{(2n\pi)}{n} \right]$$

Solution

There is a mistake in the last term.

There should be $\frac{1}{n} \cdot \left( \sin \left( \frac{\pi}{n} \right) + \sin \left( \frac{2\pi}{n} \right) + \cdots + \sin \left( \frac{n\pi}{n} \right) \right)$ instead of

$$\frac{1}{n} \cdot \left( \sin \left( \frac{\pi}{n} \right) + \sin \left( \frac{2\pi}{n} \right) + \cdots + \sin \left( \frac{2n\pi}{n} \right) \right).$$

Sum $\frac{\pi}{n} \sum_{i=1}^{n} \sin \left( \frac{\pi i}{n} \right)$ is the right Riemann sum for the integral $\int_{0}^{\pi} \sin x \, dx$.

So $\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} \sin \left( \frac{\pi i}{n} \right) = \frac{1}{\pi} \int_{0}^{\pi} \sin x \, dx = -\frac{1}{\pi} \cos x \big|_{x=0}^{x=\pi} = \frac{2}{\pi} \approx 0.6366$.

Answer: $\frac{2}{\pi} \approx 0.6366$.

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