Problem 75 - Calculus: Early Transcendentals

Question

Find the sum of the series. \( \sum_{n = 1}^{\infty} (-1)^{n-1} \frac {3^n}{n 5^n} \)

Step-by-Step Solution

Answer

\( \ln(\frac{5}{2}) \)

Step 1: Identify the Type of Series

The given series is \( \sum_{n = 1}^{\infty} (-1)^{n-1} \frac{3^n}{n 5^n} \). This is an alternating series because of the \((-1)^{n-1}\) term. Additionally, it can be expressed as an alternating geometric series with additional terms in the numerator.

Step 2: Simplify the General Term

Rewrite the general term. We have \((-1)^{n-1} \frac{3^n}{n 5^n} = (-1)^{n-1} \frac{(\frac{3}{5})^n}{n}\). This can now be recognized as an alternating series with terms \((-1)^{n-1} \frac{(\frac{3}{5})^n}{n}\).

Step 3: Use the Alternating Series Test

For the series \( \sum_{n=1}^{\infty} (-1)^{n-1} b_n \), with \( b_n = \frac{(\frac{3}{5})^n}{n} \), we apply the Alternating Series Test. Two conditions must be satisfied: \( b_n \) must be decreasing and \( \lim_{n \to \infty} b_n = 0 \).

Step 4: Verify Convergence Conditions

First, \( b_n = \frac{(\frac{3}{5})^n}{n} \) is decreasing because both \( \left( \frac{3}{5} \right)^n \) decreases as \( n \) increases and \( \frac{1}{n} \) is also decreasing. Second, \( \lim_{n \to \infty} \frac{(\frac{3}{5})^n}{n} = 0 \) since \( (\frac{3}{5})^n \to 0 \) and \( \frac{1}{n} \to 0 \). Thus, the series converges by the Alternating Series Test.

Step 5: Recognize Special Function

The given series is a special case of the logarithmic series, related to \( \ln(1 - x) = -\sum_{n=1}^{\infty} \frac{x^n}{n} \). Here, the series becomes \( -\ln(1 - (-\frac{3}{5})) \) or simply \( \ln(\frac{5}{2}) \).

Step 6: Calculate the Sum

Since the series is \( \ln(\frac{5}{2}) \), we have the sum of the series as \( \ln(\frac{5}{2}) \).

Key Concepts

Convergence

In the context of an infinite series, convergence refers to the behavior of the series as the number of terms increases indefinitely. A series converges if the sum of its terms approaches a specific value. In other words, the partial sums of the series get closer and closer to a certain number, known as the limit.

• If a series converges, it implies that adding more and more terms continues to approach this finite sum.
• If a series does not converge, it is said to diverge.

For the given series \[\sum_{n = 1}^{\infty} (-1)^{n-1} \frac {3^n}{n 5^n}\]we examine convergence by using tests such as the Alternating Series Test. The series passes the criteria of this test, establishing that it does indeed converge.
Understanding convergence not only allows us to determine the behavior of an infinite series but helps in calculating its sum, provided the series converges to a finite value.

Alternating Series Test

The Alternating Series Test is a vibrant tool in calculus for determining the convergence of an alternating series. An alternating series is one in which the signs of its terms alternate, typically following the pattern \((-1)^{n-1}\).
For instance, in our original series:

• We have \((-1)^{n-1} \frac{3^n}{n 5^n}\).
• The sequence \(b_n = \frac{3^n}{n 5^n}\) underlies the alternating term and needs to be focused on for the test.

The Alternating Series Test states that if \(b_n\) is decreasing and its limit approaches zero as \(n\) approaches infinity, then the series converges. For this test, the conditions met are:

• \(b_n\) is decreasing; since \(\left(\frac{3}{5}\right)^n\) decreases and \(\frac{1}{n}\) also decreases as \(n\) increases.
• \(\lim_{n \to \infty} b_n = 0\) which is true in this case.

This, solidly, demonstrates the series' convergence, showing how powerful the Alternating Series Test is.

Logarithmic Series

A logarithmic series is a special power series tied to the natural logarithm function. It typically takes the shape \(\ln(1-x) = -\sum_{n=1}^{\infty} \frac{x^n}{n}\).This equation forms a foundation in expanding the natural logarithmic functions into infinite series representations.

• It can compute logarithms for numbers close to 1 effectively.
• In this exercise, the series relates to \(\ln(\frac{5}{2})\), through the formula by substituting \(x = -\frac{3}{5}\).

The conversion of the series to a logarithmic form exemplifies how series are used to dive into complex functions. Recognizing this form allows us to directly equate the sum of the series, as seen with \(\ln(\frac{5}{2})\).The concept of a logarithmic series not only uncovers deep connections between algebra and calculus but also provides practical computations in mathematics, especially in number theory and numerical analysis.