Problem 46 - Calculus Early Transcendentals: Pearson New International Edition

Question

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) \(\sum_{n=1}^{\infty} \sqrt{n}\left[1-\cos \left(\frac{1}{n}\right)\right]\)

Step-by-Step Solution

Answer

(a) Converges, (b) Diverges, (c) Converges.

Step 1: Convergence Test for Series (a)

Consider the series \( \sum_{n=1}^{\infty} \sin^2\left(\frac{1}{n}\right) \). Use the limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Therefore, when \( n \) is large, \( \sin\left(\frac{1}{n}\right) \approx \frac{1}{n} \). This implies \( \sin^2\left(\frac{1}{n}\right) \approx \left(\frac{1}{n}\right)^2 = \frac{1}{n^2} \). The series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is a convergent p-series with \( p=2 > 1 \). Hence, the given series converges.

Step 2: Convergence Test for Series (b)

Consider the series \( \sum_{n=1}^{\infty} \tan\left(\frac{1}{n}\right) \). For small \( x \), \( \tan x \approx x \), so \( \tan\left(\frac{1}{n}\right) \approx \frac{1}{n} \). Therefore, the series \( \sum_{n=1}^{\infty} \frac{1}{n} \) resembles the harmonic series, which is divergent. Hence, the series is divergent.

Step 3: Convergence Test for Series (c)

Consider \( \sum_{n=1}^{\infty} \sqrt{n}\left[1-\cos\left(\frac{1}{n}\right)\right] \). Using the identity \( 1-\cos(x) \approx \frac{x^2}{2} \) for small \( x \), we approximate \( 1-\cos\left(\frac{1}{n}\right) \approx \frac{1}{2n^2} \). Therefore, \( \sqrt{n}\left[1-\cos\left(\frac{1}{n}\right)\right] \approx \sqrt{n} \cdot \frac{1}{2n^2} = \frac{1}{2n^{3/2}} \). \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \) is a p-series with \( p=3/2 > 1 \) and is convergent. Thus, the series is convergent.

Key Concepts

Convergence Tests

Convergence tests help us determine whether an infinite series converges or diverges. This is essential because knowing if a series converges means the sum of its infinite terms approaches a finite number. Various tests are available, each suited for different types of series. Here are some that you might encounter:

• Comparison Test: Compare the series in question to a known series. If the series you're comparing to converges and your series is smaller term-wise, then yours converges too.
• Ratio Test: Useful when terms of the series involve factorials or exponentials. It involves taking the limit of the ratio of consecutive terms.
• Root Test: Similar to the Ratio Test, but involves taking the nth root of terms.
• Limit Comparison Test: Involves comparing the limit of the ratio of terms from two sequences.
• Integral Test: If you can express the terms as a function that you integrate, this test can determine convergence.

The choice of test depends on the series' form. Often, you might try one or more methods to conclusively decide about the series' behavior.

P-Series

A P-Series is a special type of series that takes the form \[\sum_{n=1}^{fty} \frac{1}{n^p}\]Where \( p \) is a constant. The convergence of a P-Series depends on the value of \( p \):

• Convergent: If \( p > 1 \), the series converges. This includes notable examples like \( \sum \frac{1}{n^2} \), which converges to \( \frac{\pi^2}{6} \).
• Divergent: If \( p \leq 1 \), the series diverges. The Harmonic Series, which has \( p = 1 \), falls into this category.

Understanding P-Series is incredibly useful because they often form the backbone for comparison in other series convergence tests. By relating an unknown series to a known P-Series, we can quickly determine the unknown series' behavior.

Harmonic Series

The Harmonic Series is one of the most well-known series in mathematics, represented as:\[\sum_{n=1}^{\infty} \frac{1}{n}\]Despite its simple form and similarity to other convergent series, the Harmonic Series actually diverges. That means even though each term is getting smaller and tends to zero, their sum does not settle at a finite number. Here's some interesting further detail:

• Each term \( \frac{1}{n} \) becomes incredibly small, yet insufficiently fast to allow convergence.
• It shows that not all series that tend to zero necessarily converge.
• The divergence of the Harmonic Series can be proven by comparing it to the integral of \( \frac{1}{x} \), which diverges as \( x \to \infty \).

The divergent nature of the Harmonic Series is a paradoxical and intriguing concept for many learners, encouraging a deeper exploration of series convergence.