Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

Determine convergence or divergence for each of the series. Indicate the test you use. $$ \sum_{n=1}^{\infty} \frac{n^{2}+1}{3^{n}} $$

Step-by-Step Solution

Answer

The series converges by the Ratio Test.

Step 1: Express the Series

First, let's write the series in the form it is given: \[\sum_{n=1}^{\infty} \frac{n^2 + 1}{3^n}.\]

Step 2: Identify the Test to Use

To determine the convergence or divergence of the series, we will utilize the Ratio Test because the series has a term with a factorial-like growth (\(n^2\)) over an exponential term (\(3^n\)). The Ratio Test is useful in handling series with these types of terms.

Step 3: Apply the Ratio Test

The Ratio Test states that for a series \(\sum a_n\), consider the limit \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|.\) If \(L < 1\), the series converges. If \(L > 1\), the series diverges. If \(L = 1\), the series may converge or diverge.Calculate:\[a_n = \frac{n^2 + 1}{3^n},\quad a_{n+1} = \frac{(n+1)^2 + 1}{3^{n+1}}.\]Then:\[\left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)^2 + 1}{3^{n+1}} \cdot \frac{3^n}{n^2 + 1} = \frac{(n+1)^2 + 1}{3(n^2 + 1)}.\]

Step 4: Compute the Limit

Find the limit as \(n\) approaches infinity:\[L = \lim_{n \to \infty} \frac{(n+1)^2 + 1}{3(n^2 + 1)} = \lim_{n \to \infty} \frac{n^2 + 2n + 2}{3n^2 + 3}.\]Divide the numerator and the denominator by \(n^2\):\[L = \lim_{n \to \infty} \frac{1 + \frac{2}{n} + \frac{2}{n^2}}{3 + \frac{3}{n^2}} = \frac{1}{3}.\]

Step 5: Conclude Using the Ratio Test

Since \(L = \frac{1}{3} < 1\), by the Ratio Test, the series \(\sum_{n=1}^{\infty} \frac{n^2 + 1}{3^n}\) converges.

Key Concepts

Ratio Test

The Ratio Test is a powerful method used to determine the convergence or divergence of an infinite series. It is particularly useful when the series involves terms that grow rapidly, such as those involving factorials or exponential functions. In the Ratio Test, we examine the ratio of successive terms of the series.
For a series\( \sum a_n \), the test considers the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \), the series diverges.
• If \( L = 1 \), the test is inconclusive.

Utilizing this test simplifies handling series with exponential growth or decay, as it clearly demonstrates the behavior of the series' terms as \( n \) becomes large. This test is particularly fitting for series with both polynomial and exponential components, aiding in determining whether the exponential growth or decay outweighs the polynomial terms.

Convergence of Infinite Series

The convergence of an infinite series refers to whether the sum of its terms approaches a finite number as we include more terms in the sum. Understanding convergence is crucial in analysis and calculus, as it tells us whether a sum will stabilise at a particular value.
When dealing with a series, such as \( \sum_{n=1}^{\infty} a_n \), convergence means that the partial sums \( S_n = a_1 + a_2 + \cdots + a_n \) tend toward a specific finite limit as \( n \) becomes very large.

• A series converges if these partial sums approach a finite limit.
• A series diverges if the partial sums grow indefinitely or oscillate without approaching a limit.

Through various convergence tests, like the Ratio Test, we can analyze and classify the behavior of infinite series, helping us understand not just their sum, but also their implications in broader mathematical contexts.

Limit Calculation

Calculating limits is a fundamental procedure in determining the behavior of functions and sequences as their variables approach particular points or infinity. In the context of series convergence, limit calculations help evaluate the result of convergence tests, such as the Ratio Test.
In the solution provided, we calculated the limit \( L = \lim_{n \to \infty} \frac{(n+1)^2 + 1}{3(n^2 + 1)} \), applying algebraic manipulations to simplify the expression.

• Divide terms by the highest power of \( n \) to bring out constant terms, simplifying the calculation.
• Observe that as \( n \) tends to infinity, terms of lower degree become negligible.
• In this example, simplifying gave us \( L = \frac{1}{3} \).

This calculation step is essential in testing convergence; the simplified limit provides a concise result that allows us to apply the rules of the convergence test effectively.

Exponential Decay in Series

Exponential decay refers to how quantities decrease at a rate proportional to their current value. In the context of series, exponential decay influences how series terms behave and often dictates the overall convergence of the series.
For example, in a series like \( \sum_{n=1}^{\infty} \frac{n^2 + 1}{3^n} \), the exponential term \( 3^n \) in the denominator leads to decay, overshadowing polynomial growth from \( n^2 + 1 \).

• Exponential decay ensures terms decrease rapidly, crucial for convergence.
• Common in mathematical modeling, capturing processes where decay accelerates over time.
• In the context of this series, the decay is fast enough to ensure convergence, as verified by the Ratio Test.

Understanding this concept in series analysis helps predict where series will converge and provides clarity in many scientific applications involving natural exponential growth and decay.