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Magazine Chapter 11: Problem 23
Use comparison tests to determine whether the infinite series converge or diverge. \(\sum_{n=1}^{x} \frac{\sin ^{2}(1 / n)}{n^{2}}\)
Short Answer
Expert verified
The series converges by the Limit Comparison Test.
Step by step solution
01
Understanding the Series
The given series is \( \sum_{n=1}^{\infty} \frac{\sin^2(1/n)}{n^2} \). Our goal is to use comparison tests to determine if this series converges or diverges.
02
Simplifying the Terms
Notice that for small values of \( n \), \( \sin(1/n) \approx 1/n \). Therefore, \( \sin^2(1/n) \approx (1/n)^2 \), leading to the approximation \( \frac{\sin^2(1/n)}{n^2} \approx \frac{1/n^2}{n^2} = \frac{1}{n^4} \).
03
Use the Limit Comparison Test
Compare \( \frac{\sin^2(1/n)}{n^2} \) to \( \frac{1}{n^4} \) to see if we can conclude anything about convergence. The Limit Comparison Test states that if \( \lim_{n \to \infty} \frac{a_n}{b_n} = c \) where \( 0 < c < \infty \), then both series \( \sum a_n \) and \( \sum b_n \) either both converge or both diverge.
04
Calculate the Limit
Calculate \( \lim_{n \to \infty} \frac{\frac{\sin^2(1/n)}{n^2}}{\frac{1}{n^4}} = \lim_{n \to \infty} n^2 \sin^2(1/n) \). Since \( \sin(1/n) \approx 1/n \), it follows that \( n \sin(1/n) \approx 1 \). Therefore, \( n^2 \sin^2(1/n) \approx 1 \), giving \( \lim_{n \to \infty} n^2 \sin^2(1/n) = 1 \).
05
Conclusion via Comparison
Since \( \lim_{n \to \infty} \frac{\frac{\sin^2(1/n)}{n^2}}{\frac{1}{n^4}} = 1 \) and the series \( \sum_{n=1}^{\infty} \frac{1}{n^4} \) is known to converge (as a p-series with \( p > 1 \)), the original series \( \sum_{n=1}^{\infty} \frac{\sin^2(1/n)}{n^2} \) also converges by the Limit Comparison Test.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit Comparison Test
The Limit Comparison Test is a handy tool for determining the convergence or divergence of infinite series. To use it, we compare two series: the one we are interested in ( \( \sum a_n \) ) and a known benchmark series ( \( \sum b_n \)\ ).
- The test states that if the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} = c \) exists and is positive (\( 0 < c < \infty \)), then both series will share the same fate: they either both converge or both diverge.
- This is particularly useful because usually, one of the series ( \( \sum b_n \)\ ) is simpler and its convergence properties are known.
Series Convergence
Series convergence is about determining whether the sum of infinitely many terms results in a finite number. An infinite series \( \sum_{n=1}^{\infty} a_n \) is said to converge if the sequence of partial sums \( S_N = a_1 + a_2 + \dots + a_N \) approaches a finite limit as \( N \to \infty \).
- If the series converges, the infinite summation effectively "adds up" to a finite value.
- Conversely, if the partial sums don鈥檛 approach a limit, the series diverges, meaning it doesn鈥檛 result in a finite value.
Convergence and Divergence
Understanding convergence and divergence is key to analyzing series. Convergence signifies that the infinite sum stabilizes at a certain value. Divergence, however, means the sum doesn't settle to any finite number.
- To test for convergence, various methods are used, of which the Limit Comparison Test is one.
- Divergence occurs when terms don鈥檛 significantly diminish or cancel out as \( n \to \infty \), causing the total to grow indefinitely.
P-Series
The P-Series is a well-known series type of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a positive constant. It converges when \( p > 1 \) and diverges for \( p \leq 1 \).
- For example, the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges because \( p = 2 > 1 \).
- On the contrary, the Harmonic Series \( \sum_{n=1}^{\infty} \frac{1}{n} \) diverges since \( p = 1 \leq 1 \).
