91Ó°ÊÓ

The Integral Test is a powerful tool used to determine the convergence of infinite series. It gives us a way to compare a series to an improper integral. Here, we focus on positive decreasing functions, where the series \( \sum_{n=a}^{\infty} f(n) \) can be compared to the improper integral \( \int_{a}^{\infty} f(x) \, dx \). This principle helps us determine if a series converges or diverges.

To apply the Integral Test effectively, follow these steps:

• Ensure the function \( f(x) \) is positive, continuous, and decreasing on \([a, \, \infty)\).
• Compute the improper integral \( \int_{a}^{\infty} f(x) \, dx \).

If the integral converges, the series converges. If the integral diverges, the series does too.

In our series example \( \sum_{n=1}^{\infty} n^{-1.001} \), the function \( f(x) = x^{-1.001} \) meets these criteria, enabling us to apply the Integral Test to understand the convergence behavior.

Improper integrals are used in the context of convergence to handle infinite limits or unbounded functions. They extend the concept of a definite integral, allowing us to calculate areas under curves that stretch to infinity.

In our problem, we explore the integral \( \int_{n}^{\infty} x^{-1.001} \, dx \), a classic example of an improper integral because the upper limit extends to infinity. Calculating this involves finding the antiderivative and computing the limit as the upper bound approaches infinity.

The computation is given by:\[\int x^{-1.001} \, dx = \frac{x^{-0.001}}{-0.001} = -1000 \cdot \frac{1}{x^{0.001}}\]Applying limits, we find:\[-1000 \left[ \frac{1}{x^{0.001}} \right]_{n}^{\infty} = \frac{1000}{n^{0.001}}\]This result helps us approximate the behavior of the series, providing insight into how error can be managed through proper selection of \( n \).

Error estimation is crucial when approximating the sum of an infinite series using a finite number of terms. It ensures that our approximation is sufficiently accurate by setting a limit on the "leftover" or error term.

In the series \( \sum_{n=1}^{\infty} n^{-1.001} \), we wish to ensure a high degree of accuracy, with the error less than 5 in the ninth decimal place. We achieve this by solving for \( n \) to ensure \( \frac{1000}{n^{0.001}} < 5 \times 10^{-9} \).

To find \( n \):

• Rearrange the inequality into \( n^{0.001} > \frac{1000}{5 \times 10^{-9}} \).
• Simplify to find \( 2 \times 10^{11} < n^{0.001} \).
• Raise both sides to the power of 1000: \( 10^{11.301} < n \).

This calculation implies that more than \( 10^{11.301} \) terms are needed for the desired precision, underscoring the importance of error analysis in series approximation.