Problem 35 - Thomas Calculus

Question

Which of the series in Exercises \(17-56\) converge, and which diverge? Use any method, and give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{1-n}{n 2^{n}} $$

Step-by-Step Solution

Answer

The series converges by the Ratio Test.

Step 1: Identify the Type of Series

The given series is \(\sum_{n=1}^{\infty} \frac{1-n}{n 2^{n}}\). This is an infinite series where each term is \(a_n = \frac{1-n}{n 2^{n}}\). We need to figure out if it converges or diverges.

Step 2: Check for Convergence Using the Ratio Test

The Ratio Test is one way to determine if a series converges. For a series \(\sum a_n\), the Ratio Test suggests examining \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). If the result is less than 1, the series converges. Let's compute this for our series.

Step 3: Compute the Ratio

We have \(a_n = \frac{1-n}{n 2^{n}}\) and \(a_{n+1} = \frac{1-(n+1)}{(n+1) 2^{n+1}}\). Compute the ratio:\[\frac{a_{n+1}}{a_n} = \frac{\frac{1-(n+1)}{(n+1) 2^{n+1}}}{\frac{1-n}{n 2^{n}}} = \frac{(1-n)n 2^n}{(2^{n+1})(1-(n+1)) (n+1)}\]Simplifying:\[= \frac{n(1-n)2^n}{2^{n+1}(-(n))) (n+1)} = \frac{n(1-n)}{2(-n)(n+1)}\]

Step 4: Simplify and Analyze the Limit

Further simplification gives:\[= \frac{1-n}{-2(n+1)}\]As \(n\) approaches infinity, observe \(\lim_{n \to \infty} \left| \frac{1-n}{-2(n+1)} \right|\). Simplifying the expression as \(n\to\infty\), we focus on the leading term:\[\approx \frac{n}{2n} = \frac{1}{2}\]

Step 5: Conclusion on Convergence of the Series

Since the limit from the Ratio Test is \(\frac{1}{2} < 1\), the series \(\sum_{n=1}^{\infty} \frac{1-n}{n 2^{n}}\) converges.

Key Concepts

Infinite Series

An infinite series is essentially the sum of an infinite sequence of numbers. Imagine starting with the first number, adding the second, then the third, and continuing infinitely.
A series is represented notationally as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) denotes each term in the sequence.
Understanding whether these series add up to a finite number or not is critical in mathematics and leads us to the notion of convergence or divergence.

• Convergent Series: These series tend to a particular value as you sum more and more terms.
• Divergent Series: These series grow indefinitely or oscillate as you continue to add terms.

The series in question, \( \sum_{n=1}^{\infty} \frac{1-n}{n 2^{n}} \), is an infinite series, and it's crucial to determine if it converges to some value or not.

Ratio Test for Convergence

The Ratio Test is an essential tool used to determine the convergence of an infinite series.
This test examines the limit of the absolute value of the ratio of successive terms in the series.
The key steps involve calculating the limit:\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]Here's how the test guides us:

• If the limit is less than 1, the series converges.
• If the limit is greater than 1, the series diverges.
• If the limit equals 1, the test is inconclusive.

For the series \( \sum_{n=1}^{\infty} \frac{1-n}{n 2^{n}} \), the Ratio Test helps us find:\[ \lim_{n \to \infty} \left| \frac{1-n}{-2(n+1)} \right| = \frac{1}{2} \]Since \( \frac{1}{2} < 1 \), the series converges. This concise conclusion helps determine the nature of our series without viewing every individual term.

Limit Evaluation

Limit evaluation plays a crucial role in understanding the behavior of functions and sequences as they tend towards infinity, or a particular point of interest.
In the context of series convergence, we often simplify terms by observing the behavior as \( n \) grows larger.
For the series \( \sum_{n=1}^{\infty} \frac{1-n}{n 2^{n}} \), we assess:\[ \lim_{n \to \infty} \left| \frac{1-n}{-2(n+1)} \right| \]We focus on how the terms adjust with \( n \). As \( n \to \infty \), the dominant term \( n \) in both the numerator and denominator simplifies to:\[ \approx \frac{n}{2n} = \frac{1}{2} \]This simplification confirms that the limit of the ratio test remains \( \frac{1}{2} \), indicating convergence. Limits thereby serve as an invaluable computational tool in deciding the fate of series, allowing us, without exhaustive computation, to ascertain convergence.