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Magazine Chapter 9: Problem 46
Test for convergence or divergence. (a) \(\sum_{n=1}^{\infty} \sin ^{2}\left(\frac{1}{n}\right)\) (b) \(\sum_{n=1}^{\infty} \tan \left(\frac{1}{n}\right)\) (c) \(\sum_{n=1}^{\infty} \sqrt{n}\left[1-\cos \left(\frac{1}{n}\right)\right]\)
Short Answer
Expert verified
(a) converges; (b) diverges; (c) converges.
Step by step solution
01
Understand Convergence Tests
A series \( \sum a_n \) converges if the sequence of partial sums \( S_n = a_1 + a_2 + \cdots + a_n \) approaches a finite limit as \( n \) approaches infinity. To determine convergence or divergence, we often use tests like the Limit Comparison Test, Integral Test, or others.
02
Analyze Series (a)
Consider the series \( \sum_{n=1}^{\infty} \sin^{2}\left(\frac{1}{n}\right) \). For small values of \( x \), \( \sin(x) \approx x \), so \( \sin^{2}\left(\frac{1}{n}\right) \approx \left(\frac{1}{n}\right)^2 \). We compare \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), a known convergent p-series (with \( p=2 \)), to our series. Hence, since \( \left(\frac{1}{n}\right)^2 \) converges, by the Limit Comparison Test, \( \sum_{n=1}^{\infty} \sin^{2}\left(\frac{1}{n}\right) \) also converges.
03
Analyze Series (b)
Consider \( \sum_{n=1}^{\infty} \tan\left(\frac{1}{n}\right) \). Using a Taylor expansion, \( \tan(x) \approx x + \frac{x^3}{3} + \cdots \) for small \( x \), so \( \tan\left(\frac{1}{n}\right) \approx \frac{1}{n} \), which is a harmonic series known to diverge. Therefore, \( \sum_{n=1}^{\infty} \tan\left(\frac{1}{n}\right) \) diverges.
04
Analyze Series (c)
Consider \( \sum_{n=1}^{\infty} \sqrt{n}\left[1-\cos\left(\frac{1}{n}\right)\right] \). For small \( x \), \( 1 - \cos(x) \approx \frac{x^2}{2} \), so \( 1 - \cos\left(\frac{1}{n}\right) \approx \frac{1}{2n^2} \). Therefore, \( \sqrt{n} \left(1 - \cos\left(\frac{1}{n}\right)\right) \approx \frac{\sqrt{n}}{2n^2} = \frac{1}{2n^{3/2}} \). Since \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \) converges (a p-series with \( p > 1 \)), \( \sum_{n=1}^{\infty} \sqrt{n}\left[1-\cos\left(\frac{1}{n}\right)\right] \) converges by comparison.
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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 helpful tool for determining the convergence or divergence of a series. It allows you to compare the series in question to a known convergent or divergent series. If you find a similar series, you can infer the behavior of your original series. Here's how it works:
- Consider the series \( \sum a_n \) that you want to test for convergence.
- Identify another series \( \sum b_n \) that is known to converge or diverge.
- Calculate the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \).
- If this limit is a positive finite number, then both series \( \sum a_n \) and \( \sum b_n \) behave the same way鈥攊f one converges, so does the other.
p-series convergence
A p-series is any series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a constant. The behavior of p-series is well-known:
- If \( p > 1 \), the p-series converges.
- If \( p \leq 1 \), the p-series diverges.
Taylor expansion
Taylor expansion is a mathematical technique to approximate functions using polynomials. It helps in approximating the behavior of a function near a certain point.For instance, for small values of \( x \), the expansion of \( \tan(x) \) is:\[ \tan(x) \approx x + \frac{x^3}{3} + \cdots \]This simplicity is very helpful when checking series for convergence, as it allows a function like \( \tan\left(\frac{1}{n}\right) \) to resemble \( \frac{1}{n} \), suggesting a comparison to the harmonic series.In the exercise, this expansion was critical to deciding that \( \sum_{n=1}^{\infty} \tan \left(\frac{1}{n}\right) \) diverges, as it behaves similarly to the harmonic series.
Harmonic series
The harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \) is a classic example of a divergent series. Although the terms \( \frac{1}{n} \) become very small as \( n \) increases, the sum grows without bound as more terms are added.This series is pivotal when using comparison techniques, as it serves as a benchmark for divergence. Because it's simple to compare against, it was used in the Taylor expansion context when analyzing \( \sum_{n=1}^{\infty} \tan \left(\frac{1}{n}\right) \). Since \( \tan\left(\frac{1}{n}\right) \approx \frac{1}{n} \), it can be concluded that this new series also diverges due to its resemblance to the harmonic series.
Integral Test
The Integral Test is another method to determine the convergence of a series. You can approximate the sum of an infinite series by relating it to an improper integral.Here's how it works:
- Consider the series \( \sum a_n \). Find a function \( f(x) \) such that \( f(n) = a_n \) for all integers \( n \).
- If \( f(x) \) is continuous, positive, and decreasing for \( x \geq N \) (where \( N \) is some positive integer), then compute the improper integral \( \int_N^\infty f(x)\, dx \).
- If the integral converges, so does the series \( \sum a_n \). If the integral diverges, so does the series.
