Question

\(\lim _{n \rightarrow \infty} \frac{1}{n}\left(\sin ^{2} \frac{\pi}{2 n}+\sin ^{2} \frac{2 \pi}{2 n}+\ldots . . .+\sin ^{2} \frac{n \pi}{2 n}\right)\) is equal to (1) \(\frac{2}{\pi}\) (2) \(\frac{\pi}{2}\) (3) \(\frac{1}{2}\) (4) 1

Short answer

The limit is \(\frac{1}{2}\).\)

Step-by-step solution

- Understand the Limit Expression

Consider the limit expression \(\lim_{{n \to \infty}} \frac{1}{n}\bigg(\sin^2 \frac{\pi}{2n} + \sin^2 \frac{2\pi}{2n} + \ldots + \sin^2 \frac{n\pi}{2n}\bigg)\). Since the argument of the sine function is small for large n, we can use the approximation that \(\sin x \approx x for \small x.\)

- Apply Small Angle Approximation

For large n, using \(\sin x \approx x\), the expression becomes \[\frac{1}{n} \sum_{k=1}^{n} \left(\sin^2 \frac{k\pi}{2n}\right) \approx \frac{1}{n} \sum_{k=1}^{n} \left(\frac{k\pi}{2n}\right)^2.\]

- Simplify the Summation Using Approximation

Rewrite the sum \[\frac{1}{n} \sum_{k=1}^{n} \left(\frac{k\pi}{2n}\right)^2 = \frac{\pi^2}{4n^3} \sum_{k=1}^{n} k^2.\] The sum of squares formula is given by \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}.\)

- Insert and Evaluate the Series

Substituting the formula for the sum of squares, we have \[\frac{\pi^2}{4n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{\pi^2}{4n^3} \cdot \frac{2n^3 + 3n^2 + n}{6}.\]

- Simplify and Take the Limit

Simplify the expression: \[\frac{\pi^2}{4n^3} \cdot \frac{2n^3 + 3n^2 + n}{6} = \frac{\pi^2}{4n^3} \cdot \frac{2n^3 + O(n^2)}{6} \approx \frac{\pi^2}{4n^3} \cdot \frac{2n^3}{6} = \frac{\pi^2 \cdot 2}{24} = \frac{\pi^2}{12}.\] Taking the limit as \(n \to \infty\), we obtain \[\lim_{{n \to \infty}} \frac{\pi^2}{12} = \frac{1}{2}.\]

Key concepts

Limit Evaluation

Limit evaluation is a core concept needed for solving many mathematical problems, especially those in calculus. When you evaluate a limit, you determine the value that a function approaches as its input approaches a certain point. For the given problem, the limit of a summation involving sine squared terms as n approaches infinity must be evaluated. This involves a few critical steps:

• Recognize that the sine function's arguments become very small for large n.
• Use approximations to simplify the expression.
• Apply mathematical formulas to handle summations.

By following these steps, complex expressions can be evaluated accurately, even when they involve infinite series.

Small Angle Approximation

Small angle approximation is a useful tool in trigonometry. It simplifies calculations by approximating trigonometric functions for small values of their arguments. For the sine function, the approximation \(\sin(x) \approx x\) holds true when x is close to zero. This simplification is pivotal in the given exercise, especially since \(\sin^2(x)\) can be approximated as \[ x^2 \].
This approximation allows us to transform the limit expression from one involving multiple sine terms to a simpler form involving squares of fractions.

For instance:

• Transform \(\sin^2 \left( \frac{k \pi}{2n} \right)\) to \(\left( \frac{k \pi}{2n}\right)^2\).

This simplification significantly reduces complexity and makes it easier to apply summation formulas subsequently.

Summation Formulas

Summation formulas are crucial for solving series and understanding patterns within numbers. They allow us to sum up sequences of numbers efficiently. For the given problem, the formula for the sum of squares is particularly useful: \[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \].
This formula helps in transforming a lengthy summation into a manageable expression. Here's how it applies in the exercise:

• Replace the sum of the series of squares in the limit expression.
• Simplify the resulting expression to evaluate the limit as n approaches infinity.

Summation formulas thus provide a structured approach to handle and simplify complex sums, leading to an accurate evaluation of limits in problems similar to this.