Math Problem Statement

Determine whether the series Σ sin(1 / n^2) (n=1 to ∞) converges or diverges. Justify your answer.

Solution

We are tasked to determine whether the series

n=1sin(1n2)\sum_{n=1}^\infty \sin\left(\frac{1}{n^2}\right)

converges or diverges and to justify the conclusion.


Analysis:

  1. Behavior of the Term:

    • For large nn, 1n2\frac{1}{n^2} approaches 0.
    • Using the Taylor expansion for sin(x)\sin(x) around 0: \sin(x) \approx x \quad \text{(for small x)}.
    • Hence, for large nn, sin(1n2)1n2.\sin\left(\frac{1}{n^2}\right) \approx \frac{1}{n^2}.
  2. Comparison to a Known Series:

    • The series n=11n2\sum_{n=1}^\infty \frac{1}{n^2} is a p-series with p=2>1p = 2 > 1, and it converges.
    • Since sin(1n2)1n2\sin\left(\frac{1}{n^2}\right) \sim \frac{1}{n^2}, we suspect n=1sin(1n2)\sum_{n=1}^\infty \sin\left(\frac{1}{n^2}\right) converges.
  3. Limit Comparison Test:

    • Let an=sin(1n2)a_n = \sin\left(\frac{1}{n^2}\right) and bn=1n2b_n = \frac{1}{n^2}.
    • Compute: limnanbn=limnsin(1n2)1n2.\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\sin\left(\frac{1}{n^2}\right)}{\frac{1}{n^2}}.
    • For small xx, sin(x)x\sin(x) \approx x, so: sin(1n2)1n21as n.\frac{\sin\left(\frac{1}{n^2}\right)}{\frac{1}{n^2}} \to 1 \quad \text{as } n \to \infty.
    • Since the limit is a positive finite constant (11), and bn=1n2\sum b_n = \sum \frac{1}{n^2} converges, the Limit Comparison Test implies an\sum a_n also converges.

Conclusion:

The series n=1sin(1n2)\sum_{n=1}^\infty \sin\left(\frac{1}{n^2}\right) converges. This conclusion is justified by approximating the terms and applying the Limit Comparison Test.


Do you want a detailed explanation of the Limit Comparison Test or the derivation of the Taylor expansion? Here are related questions to expand on this:

  1. How does the Taylor series approximation for sin(x)\sin(x) work, and what are its higher-order terms?
  2. Why does the series 1n2\sum \frac{1}{n^2} converge, and how does this relate to p-series?
  3. Can the Direct Comparison Test be applied here instead of the Limit Comparison Test?
  4. How does the behavior of sin(x)\sin(x) influence series convergence for different powers of nn?
  5. What happens if we replace sin(1n2)\sin\left(\frac{1}{n^2}\right) with sin(1n)\sin\left(\frac{1}{n}\right)?

Tip: When approximating terms in series, always check the validity of using asymptotic approximations and whether a formal convergence test is necessary.

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Math Problem Analysis

Mathematical Concepts

Infinite series
Convergence and divergence
Limit comparison test
Taylor series expansion

Formulas

sin(x) ≈ x (for small x)
Limit comparison test: lim (a_n / b_n) as n -> ∞

Theorems

Limit Comparison Test
p-series convergence theorem

Suitable Grade Level

Undergraduate - Calculus II