Question

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

Short answer

The series converges by the Limit Comparison Test with \( \sum_{n=1}^{\infty} \frac{1}{4^n} \).

Step-by-step solution

Identify the Series

The given series is \( \sum_{n=1}^{\infty} \frac{1}{4^{n}-n^{2}} \). We are tasked to use the Limit Comparison Test to determine its convergence.

Choose a Comparison Series

To apply the Limit Comparison Test, we need a series whose convergence is already known. We choose the geometric series \( \sum_{n=1}^{\infty} \frac{1}{4^n} \), which converges because \( \frac{1}{4} < 1 \).

Formulate the Limit for the Comparison Test

We calculate the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \), where \( a_n = \frac{1}{4^n - n^2} \) and \( b_n = \frac{1}{4^n} \). Therefore, the limit becomes: \[ \lim_{n \to \infty} \frac{\frac{1}{4^n-n^2}}{\frac{1}{4^n}} = \lim_{n \to \infty} \frac{4^n}{4^n-n^2}. \]

Evaluate the Limit

Simplify the expression: \[ L = \lim_{n \to \infty} \frac{4^n}{4^n - n^2}. \] Divide the numerator and denominator by \( 4^n \): \[ L = \lim_{n \to \infty} \frac{1}{1 - \frac{n^2}{4^n}}. \] As \( n \to \infty \), \( \frac{n^2}{4^n} \to 0 \), so \( L = 1 \).

Apply the Limit Comparison Test

Since the limit \( L = 1 \) is positive and finite, and the comparison series \( \sum_{n=1}^{\infty} \frac{1}{4^n} \) converges, by the Limit Comparison Test, the series \( \sum_{n=1}^{\infty} \frac{1}{4^n - n^2} \) also converges.

Key concepts

Understanding Geometric Series

A geometric series is a series of numbers in which each term after the first is obtained by multiplying the previous one by a fixed, non-zero number called the common ratio. This type of series can be expressed as \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term and \( r \) is the common ratio.
Understanding geometric series is essential as they allow us to determine the behavior of more complex series using various convergence tests.

If \( |r| < 1 \), the geometric series converges to \( \frac{a}{1-r} \). This is because the terms become smaller and closer to zero, allowing the sum to settle towards a finite number.
For example, the series \( \sum_{n=0}^{\infty} \frac{1}{4^n} \) is a geometric series with \( a = 1 \) and \( r = \frac{1}{4} \), which converges since \( \frac{1}{4} < 1 \).

This convergence property is crucial when using the Limit Comparison Test to evaluate the behavior of series that resemble geometric ones.

Convergence of Series

The convergence of a series is a fundamental concept in mathematical analysis. When we talk about the convergence of a series, we mean that the sum of its terms approaches a specific finite limit as the number of terms increases to infinity.
The opposite is divergence, where the sum fails to settle at a particular value and continues growing or fluctuating.

For a series like \( \sum_{n=1}^{\infty} \frac{1}{4^n} \), convergence is straightforward because it is a geometric series with a common ratio less than 1.
The convergence of more complex series, such as \( \sum_{n=1}^{\infty} \frac{1}{4^n - n^2} \), often requires additional tests like the Limit Comparison Test.
This test helps us determine the convergence of a series by comparing it with another series whose convergence behavior is already known.

By understanding whether a series converges, we can ascertain its potential influence or result in larger mathematical contexts, such as calculus or physics applications.

Calculation of Limits in Series Comparison

Calculating limits plays a critical role in evaluating the behavior of series, especially when using comparison tests like the Limit Comparison Test.
To apply this test to determine whether a series converges, we compute the limit of the ratio of its terms against those of a known convergent series.

Let's look at the calculation of the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \). For the given series \( a_n = \frac{1}{4^n - n^2} \) compared with \( b_n = \frac{1}{4^n} \), the limit simplifies to:

• \( L = \lim_{n \to \infty} \frac{4^n}{4^n - n^2} \)
• By dividing both the numerator and the denominator by \( 4^n \), we get \( L = \lim_{n \to \infty} \frac{1}{1 - \frac{n^2}{4^n}} \).

This manipulation shows that as \( n \to \infty \), the expression \( \frac{n^2}{4^n} \to 0 \), resulting in the limit \( L = 1 \).

With \( L = 1 \), which is finite and positive, the Limit Comparison Test confirms that the original complex series converges just like the geometric comparison series. Understanding these steps in limit calculation is vital for successfully applying the Limit Comparison Test to various series.