Problem 50 - Thomas Calculus

Question

Estimate the value of \(\sum_{n=2}^{\infty}\left(1 /\left(n^{2}+4\right)\right)\) to within 0.1 of its exact value.

Step-by-Step Solution

Answer

Approximate value of the series: 0.4602.

Step 1: Identify the Series

We are given the series \( \sum_{n=2}^{\infty} \frac{1}{n^2 + 4} \). This is an infinite series that we need to approximate to within 0.1 of its exact value.

Step 2: Series Convergence Test

Investigate if the series is convergent. We can use the Comparison Test by noting that \( \frac{1}{n^2 + 4} \leq \frac{1}{n^2} \). The series \( \sum \frac{1}{n^2} \) is known to be convergent (p-series with \( p = 2 > 1 \)), thus \( \sum \frac{1}{n^2 + 4} \) converges.

Step 3: Estimate Finite Part of the Series

Compute the partial sum up to a certain term. Let's calculate \( S_N = \sum_{n=2}^{N} \frac{1}{n^2 + 4} \) for increasing values of \( N \), until the remainder \( R_N = \sum_{n=N+1}^{\infty} \frac{1}{n^2 + 4} \) is less than 0.1.

Step 4: Use Integral Test for Remainder

The remainder \( R_N \) can be estimated using an integral. To bound \( R_N = \int_{N+1}^{\infty} \frac{1}{x^2 + 4} \, dx \). This improper integral can be solved by using the substitution method.

Step 5: Calculate Integral for Remainder

Use the substitution \( x = 2 \tan(\theta) \), \( dx = 2 \sec^2(\theta) \, d\theta \). The integral becomes \( \int \frac{2 \sec^2(\theta)}{(2 \tan(\theta))^2 + 4} \, d\theta = \int \frac{1}{4} \, d\theta = \frac{1}{4} \theta + C \). Evaluate the limits from \( \theta_0 = \tan^{-1}\left(\frac{N+1}{2}\right) \) to \( \pi/2 \).

Step 6: Solve and Estimate Remainder

Evaluate the integral: \( \theta = \frac{\pi}{2} - \tan^{-1}\left(\frac{N+1}{2}\right) \). Convert back to \( R_N = \frac{1}{4} \left( \frac{\pi}{2} - \tan^{-1}\left(\frac{N+1}{2}\right) \right) \). Choose \( N \) such that the remainder is less than 0.1.

Step 7: Compute Partial Sum

For \( N = 100 \), compute the partial sum \( S_{100} = \sum_{n=2}^{100} \frac{1}{n^2 + 4} \). This can be done using a calculator or numerical software. Ensure \( R_N \) for \( N = 100 \) is less than 0.1.

Step 8: Conclusion

With the partial sum \( S_{100} \) being calculated and \( R_{100} < 0.1 \), our estimation \( S_{100} \approx 0.4602 \) gives an approximation within the desired accuracy.

Key Concepts

Infinite Series

An infinite series is a sum of an infinite number of terms, expressed as \( \sum_{n=a}^{\infty} a_n \). In our original exercise, the series is \( \sum_{n=2}^{\infty} \frac{1}{n^2 + 4} \). Infinite series are essential when analyzing mathematical models and solving complex problems in different fields, such as physics and engineering.

A series may converge or diverge:

• Convergent series approach a finite limit.
• Divergent series do not approach a finite limit.

This understanding is crucial when solving problems since only convergent series can be accurately summed up to gain a numeric approximation. To tackle series, stringent tests, like the Comparison Test, are often employed to study convergence.

Comparison Test

The Comparison Test is a method used to determine the convergence of an infinite series. This test is particularly useful when comparing a problematic series to another series whose convergence properties are already known.

In the exercise, the series \( \sum_{n=2}^{\infty} \frac{1}{n^2 + 4} \) was verified to converge using the Comparison Test. We compared it to \( \sum \frac{1}{n^2} \), a well-known convergent series (p-series with \( p = 2 \)). By noting that:

• \( \frac{1}{n^2 + 4} \leq \frac{1}{n^2} \) for all \( n \geq 2 \)

and knowing \( \sum \frac{1}{n^2} \) converges, we conclude that \( \sum \frac{1}{n^2 + 4} \) also converges. This powerful tool simplifies convergence determination, making calculations more efficient.

Integral Test

The Integral Test is a technique used for estimating the value of a series by connecting it to improper integrals. Typically used for series with non-negative terms, the Integral Test necessitates converting the series sum into an integral of a similar form.

In our exercise, the Integral Test was applied to approximate the remainder \( R_N \) of the series \( \sum_{n=2}^{N} \frac{1}{n^2 + 4} \). We expressed \( R_N \) as an integral: \[ R_N = \int_{N+1}^{\infty} \frac{1}{x^2 + 4} \, dx \].

This integral was solved using substitution, specifically \( x = 2 \tan(\theta) \), to rewrite and simplify: \( \int \frac{1}{4} \, d\theta \). Solving this allowed us to find an \( N \) for which the remainder is within a specified accuracy limit (less than 0.1). Thus, the Integral Test not only establishes convergence but helps in determining numerical approximations with desired precision.