Question

Use the limit comparison test to determine whether each of the following series converges or diverges. $$ \sum_{n=1}^{\infty}\left(\frac{1}{n}-\sin \left(\frac{1}{n}\right)\right) $$

Short answer

The series converges.

Step-by-step solution

Identify the Comparison Series

We begin by examining the term \(a_n = \frac{1}{n} - \sin\left(\frac{1}{n}\right)\). For large \(n\), \(\sin\left(\frac{1}{n}\right) \approx \frac{1}{n}\) because \(\sin(x) \approx x\) when \(x\) is close to zero. Thus, for large \(n\), \(a_n \approx 0\). We will use \(b_n = \frac{1}{n^2}\) as the comparison series based on this simplification.

Compute the Limit of the Ratio of Terms

We now find the limit \(L = \lim_{n \to \infty} \frac{a_n}{b_n}\). This becomes:\[L = \lim_{n \to \infty} \frac{\frac{1}{n} - \sin\left(\frac{1}{n}\right)}{\frac{1}{n^2}} = \lim_{n \to \infty} n^2 \left(\frac{1}{n} - \sin\left(\frac{1}{n}\right)\right)\]This can be simplified as:\[L = \lim_{n \to \infty} \left(n - n^2 \cdot \sin\left(\frac{1}{n}\right)\right)\]

Simplify the Limit Expression

We simplify the right side of the expression:\[L = \lim_{n \to \infty} \left(n - n \cdot n \cdot \sin\left(\frac{1}{n}\right)\right)\]Using the approximation \(\sin\left(\frac{1}{n}\right) \approx \frac{1}{n}\), we get:\[L = \lim_{n \to \infty} \left(n - n \cdot n \cdot \frac{1}{n}\right)\]\[L = \lim_{n \to \infty} (n - n) = 0\]

Apply the Limit Comparison Test Conclusion

The limit \(L = 0\) indicates that \(\lim_{n \to \infty} \frac{a_n}{b_n} = 0\), which means \(a_n\) behaves similar to \(b_n\) for large \(n\).\ The series \(\sum \frac{1}{n^2}\) is known to converge by the p-series test (\(p = 2 > 1\)).\ Therefore, by the limit comparison test, the given series \(\sum \left( \frac{1}{n} - \sin\left(\frac{1}{n}\right) \right)\) also converges.

Key concepts

Convergence of Series

Convergence in a series occurs when the sum of its terms approaches a specific value, rather than growing indefinitely. In mathematics, determining whether a series converges is essential for understanding its overall behavior, especially when dealing with infinite series.

When you have a series, like \[ \sum_{n=1}^{\infty}a_n \]analyzing its individual terms, \(a_n\), helps determine if the series converges. If the terms \(a_n\) do not approach zero, the series will definitely diverge.

However, even if they do approach zero, further examination is needed to confirm convergence. Techniques such as the limit comparison test come in handy here, allowing us to compare the series in question with a known converging series to draw conclusions about convergence.

P-Series Test

The p-series test is a simple yet powerful tool in determining the convergence of a series. A p-series is defined as:
\[ \sum_{n=1}^{\infty} \frac{1}{n^p}\]
where \(p\) is a real number.- **Convergence**: The series converges if \(p > 1\).- **Divergence**: The series diverges if \(p \leq 1\).

In the original problem, the comparison series was \( \sum \frac{1}{n^2} \). This is a typical p-series with \(p = 2\), which is greater than 1, indicating convergence. By comparing the given series with this known converging p-series, we can confidently conclude the behavior of the original series using the limit comparison test.

Sin Approximation

The sine function, \(\sin(x)\), is often approximated for small values of \(x\). One popular approximation is \(\sin(x) \approx x\), especially when \(x\) is approaching zero.

In the original exercise, this approximation is crucial. When looking at the expression \(\sin\left(\frac{1}{n}\right)\) for large \(n\), \(\frac{1}{n}\) tends towards zero. This means \(\sin\left(\frac{1}{n}\right) \approx \frac{1}{n}\).

Utilizing this approximation simplifies the analysis of the series \(\frac{1}{n} - \sin\left(\frac{1}{n}\right)\), allowing for a more straightforward application of the limit comparison test by making the behavior of the series terms clearer.

Infinite Series

An infinite series is simply a sum of infinitely many terms. It's denoted as:\[ \sum_{n=1}^{\infty}a_n \]
Infinite series are foundational in mathematical analysis, and they appear in various fields such as calculus and complex analysis.

Assessing whether such a series converges or diverges requires careful examination of its terms. If an infinite series converges, the sum of its terms approaches a finite value, despite the infinite nature of the series.

• **Convergent Series**: The terms add up to a finite limit.
• **Divergent Series**: The sum grows without bounds or oscillates.

Study aids like the limit comparison test and the p-series test provide methods to determine the convergence or divergence of infinite series, making complex problems more approachable.