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In Mathematics / College | 2025-07-03

Use the Comparison Test to determine whether the series converges.

[tex]\sum_{n=1}^{\infty} \frac{\sin ^2(n)}{n^2}[/tex]

Identify [tex]b_n[/tex], the series to be used for comparison, and identify the relationship between [tex]a_n[/tex] and [tex]b_n[/tex].
[tex]\begin{array}{l}
a_n=\frac{\sin ^2(n)}{n^2} \
b_n=\square \
a_n ? \vee b_n
\end{array}[/tex]

Since [tex]\sum b_n[/tex] is [$\square$] --Select--and [tex]a_n[/tex] [$\square$] ? [tex]b_n[/tex] for all [tex]n[/tex], [$\square$] --Select--.

Asked by frayca13

Answer (2)

Choose b n ​ = n 2 1 ​ as the comparison series.
Establish the relationship a n ​ = n 2 s i n 2 ( n ) ​ ≤ n 2 1 ​ = b n ​ .
Recognize that ∑ b n ​ = ∑ n 2 1 ​ is a convergent p-series with 1"> p = 2 > 1 .
Conclude that ∑ a n ​ = ∑ n 2 s i n 2 ( n ) ​ converges by the Comparison Test, since a n ​ ≤ b n ​ and ∑ b n ​ converges. co n v er g es ​

Explanation

Problem Analysis We are given the series ∑ n = 1 ∞ ​ n 2 s i n 2 ( n ) ​ and we want to use the Comparison Test to determine if it converges. The Comparison Test states that if 0 ≤ a n ​ ≤ b n ​ for all n , and ∑ b n ​ converges, then ∑ a n ​ also converges.

Finding a Comparison Series We need to find a suitable series b n ​ to compare with a n ​ = n 2 s i n 2 ( n ) ​ . Since 0 ≤ sin 2 ( n ) ≤ 1 for all n , we have 0 ≤ n 2 s i n 2 ( n ) ​ ≤ n 2 1 ​ . Therefore, we can choose b n ​ = n 2 1 ​ .

Convergence of the Comparison Series Now we need to determine if the series ∑ n = 1 ∞ ​ b n ​ = ∑ n = 1 ∞ ​ n 2 1 ​ converges. This is a p-series with p = 2 . A p-series ∑ n = 1 ∞ ​ n p 1 ​ converges if 1"> p > 1 and diverges if p ≤ 1 . Since 1"> p = 2 > 1 , the series ∑ n = 1 ∞ ​ n 2 1 ​ converges.

Applying the Comparison Test Since 0 ≤ a n ​ ≤ b n ​ for all n , and ∑ n = 1 ∞ ​ b n ​ converges, by the Comparison Test, the series ∑ n = 1 ∞ ​ a n ​ = ∑ n = 1 ∞ ​ n 2 s i n 2 ( n ) ​ also converges.

Final Answer Therefore, we have b n ​ = n 2 1 ​ and a n ​ ≤ b n ​ . Since ∑ b n ​ is convergent and a n ​ ≤ b n ​ for all n , ∑ a n ​ is convergent.


Examples
The Comparison Test is useful in many fields, such as physics and engineering, where we often encounter infinite series when modeling various phenomena. For example, when analyzing the stability of a structure or the convergence of an iterative process, we might use the Comparison Test to determine whether a series converges, which can tell us whether the structure is stable or the process converges to a meaningful solution. Understanding convergence is crucial for making accurate predictions and designing reliable systems.

Answered by GinnyAnswer | 2025-07-03

The series ∑ n = 1 ∞ ​ n 2 s i n 2 ( n ) ​ converges by the Comparison Test, using b n ​ = n 2 1 ​ as a comparison. Since a n ​ ≤ b n ​ and the series ∑ b n ​ converges, we conclude that ∑ a n ​ converges. This establishes that the original series converges as well.
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Answered by Anonymous | 2025-07-04