Problem 26 - Calculus: Early Transcendentals

Question

Test the series for convergence or divergence. \( \displaystyle \sum_{n = 1}^{\infty} \frac {n^2 + 1}{5^n} \)

Step-by-Step Solution

Answer

The series converges by the Ratio Test.

Step 1: Identify the Series Type

The given series is \( \sum_{n=1}^{\infty} \frac{n^2+1}{5^n} \). This series can be categorized as a geometric series if it can be simplified into the form \( ar^n \), where \( |r| < 1 \) indicates convergence. However, the presence of \( n^2 + 1 \) suggests it's not a simple geometric series, indicating a test for absolute convergence might be appropriate.

Step 2: Apply the Ratio Test

The Ratio Test is useful here. Consider the series \( a_n = \frac{n^2 + 1}{5^n} \). We need to find \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Calculate: \(\frac{a_{n+1}}{a_n} = \frac{(n+1)^2 + 1}{5^{n+1}} \times \frac{5^n}{n^2 + 1} = \frac{(n+1)^2 + 1}{5(n^2 + 1)} \).

Step 3: Simplify the Ratio Test Expression

Simplify \( \frac{(n+1)^2 + 1}{5(n^2 + 1)} \) to find the limit:\[ \lim_{n \to \infty} \frac{(n^2 + 2n + 2)}{5(n^2 + 1)} = \lim_{n \to \infty} \frac{n^2 + 2n + 2}{5n^2 + 5} \]. Dividing numerator and denominator by \( n^2 \), this becomes\[ \lim_{n \to \infty} \frac{1 + \frac{2}{n} + \frac{2}{n^2}}{5 + \frac{5}{n}} = \frac{1}{5} \].

Step 4: Conclude with the Ratio Test

The Ratio Test yields \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{1}{5} \). Since \( \frac{1}{5} < 1 \), the series \( \sum_{n=1}^{\infty} \frac{n^2 + 1}{5^n} \) is convergent by the Ratio Test.

Key Concepts

Ratio Test

The Ratio Test is a handy tool when determining the convergence of series, especially when factorials or exponential terms are involved.
It's about comparing the ratio of successive terms in a sequence. To apply the Ratio Test to a series \( \sum a_n \), you typically start with finding the limit:\[L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]If \( L < 1 \), the series is absolutely convergent; if \( L > 1 \), the series is divergent. When \( L = 1 \), the test is inconclusive. In the original problem, the series considered was \( \sum_{n=1}^{\infty} \frac{n^2+1}{5^n} \).
Upon applying the Ratio Test:

• Calculate \( a_{n+1} \) and \( a_n \).
• Find \( \frac{a_{n+1}}{a_n} \).
• Simplify and take the limit as \( n \to \infty \).

The calculations indicated that \( L = \frac{1}{5} \). Since it's less than 1, the series converges. The Ratio Test gives a clear criterion, making it valuable in problems where numerator and denominator grow at balanced rates.

Geometric Series

Understanding geometric series is foundational for grasping more complex series. A geometric series takes the form:\[\sum_{n=0}^{\infty} ar^n\]where \( a \) is the first term and \( r \) is the common ratio.
A geometric series converges if the absolute value of the common ratio \(|r|\) is less than 1. It diverges if \(|r| \geq 1\).In the original problem, we needed to determine if the series \( \sum_{n=1}^{\infty} \frac{n^2+1}{5^n} \) could be viewed as a geometric series. Given the \( n^2 + 1 \) term, it disrupted the fundamental geometric form, indicating that while the basic form didn't match, understanding convergence in terms of ratios remained relevant.
Despite not being a strict geometric sequence, recognizing the template helps isolate how terms like \( 5^n \) influence overall convergence in a broader framework.

Absolute Convergence

Absolute convergence extends upon the concept of basic convergence, offering a firmer grasp on series' behavior. A series \( \sum a_n \) is considered absolutely convergent if \( \sum |a_n| \) converges. This is a stricter condition than regular convergence. If a series is absolutely convergent, it is convergent; however, a convergent series need not be absolutely convergent.Why does this matter? Absolute convergence ensures stability in series, which is crucial when dealing with rearrangements or complex expressions. Practically, assessing absolute convergence might involve using tests like the

• Ratio Test
• Root Test
• Comparison Test

Returning to the original exercise, the series \( \sum_{n=1}^{\infty} \frac{n^2+1}{5^n} \) was treated with the Ratio Test, suggesting convergence. Indeed, since the Ratio Test revealed \( L = \frac{1}{5} \), the series was absolutely convergent.
Understanding this not just establishes the series' sum behavior but also its robustness across analyzing scenarios.