Question

Use the Direct Comparison Test to determine the convergence of the given series; state what series is used for comparison. $$\sum_{n=1}^{\infty} \frac{1}{4^{n}+n^{2}-n}$$

Short answer

The series converges using the comparison with the geometric series \(\frac{1}{4^n}\).

Step-by-step solution

Identify the Given Series

The series given is \( \sum_{n=1}^{\infty} \frac{1}{4^{n} + n^{2} - n} \). We need to determine if this series converges or diverges using the Direct Comparison Test.

Propose a Comparison Series

A suitable comparison series should be similar but simpler. Notice that \(4^n\) dominates \(n^2 - n\) as \(n\) becomes large, so we'll compare with \( \sum_{n=1}^{\infty} \frac{1}{4^n} \). This is a geometric series with ratio \( \frac{1}{4} \), which is less than 1.

Check for Comparison Conditions

For \(n \geq 1\), we have \(4^n > n^2 - n\), so \(4^n + n^2 - n > 4^n\). Thus, \(\frac{1}{4^n + n^2 - n} < \frac{1}{4^n}\). This satisfies the comparison condition \(0 \leq a_n \leq b_n\).

Determine the Convergence of the Comparison Series

The series \( \sum_{n=1}^{\infty} \frac{1}{4^n} \) is a geometric series with common ratio \( \frac{1}{4} \), which is less than 1. This means the series converges.

Apply the Direct Comparison Test Conclusion

Since \( \frac{1}{4^n + n^2 - n} < \frac{1}{4^n} \) for all \(n \geq 1\) and the series \( \sum_{n=1}^{\infty} \frac{1}{4^n} \) converges, the Direct Comparison Test implies that \( \sum_{n=1}^{\infty} \frac{1}{4^n + n^2 - n} \) also converges.

Key concepts

Direct Comparison Test

The Direct Comparison Test is a powerful technique used in the analysis of infinite series to determine their convergence or divergence. The basic idea is to compare a series of interest with another series that is easier to understand, often a simpler geometric or p-series.To use this test, follow these steps:

• Identify the series you want to test for convergence.
• Find a simpler, well-understood series for comparison.
• Ensure that the conditions for the comparison are satisfied—this means establishing that every term in your given series is less than or equal to the corresponding term in the comparison series.

For instance, if you have a series \( \sum_{n=1}^{\infty} a_n \) and you compare it to \( \sum_{n=1}^{\infty} b_n \), where \( a_n \leq b_n \) for all \( n \), and if the series \( \sum_{n=1}^{\infty} b_n \) is known to converge, then \( \sum_{n=1}^{\infty} a_n \) must also converge.In our exercise, the given series is \( \sum_{n=1}^{\infty} \frac{1}{4^n + n^2 - n} \). We used \( \sum_{n=1}^{\infty} \frac{1}{4^n} \) as the comparison series, showing that the original series converges.

Geometric Series

A geometric series is one where each term is a constant multiple of the previous term. This makes analyzing their convergence straightforward. The general form of a geometric series is:\[ \sum_{n=0}^{\infty} ar^n \]where \( a \) is the first term and \( r \) is the common ratio.When dealing with convergence, geometric series have specific rules:

• If \( |r| < 1 \), the series converges.
• If \( |r| \geq 1 \), the series diverges.

For a converging series, the sum is given by the formula:\[ \frac{a}{1-r} \]In our exercise, the comparison series \( \sum_{n=1}^{\infty} \frac{1}{4^n} \) is geometric with a ratio of \( \frac{1}{4} \), which is less than 1, hence it converges.

Series Convergence Analysis

Analyzing the convergence of a series involves determining whether the series sums to a finite number or not. This is crucial because it helps in understanding the behavior and properties of functions represented as series.There are different tests for series convergence:

• Direct Comparison Test: Useful when you can easily compare your series with a known convergent or divergent series.
• Geometric Series Test: Applicable directly to geometric series where you check the common ratio.
• Other tests: Such as the Ratio Test, Root Test, and Integral Test can also be applied depending on the series structure.

Our analysis using the Direct Comparison Test with a known geometric series was successful in proving the convergence of \( \sum_{n=1}^{\infty} \frac{1}{4^n+n^2-n} \). This process showcases how our understanding of simpler series can manage and solve more complicated ones. So, mastering series convergence analysis opens many doors in the world of mathematics!