Problem 59 - Thomas Calculus

Question

Which series in Exercises \(53-76\) converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. $$\sum_{n=0}^{\infty} e^{-2 n}$$

Step-by-Step Solution

Answer

The series converges, and its sum is approximately 1.156.

Step 1: Identify the Series

The series given is \(\sum_{n=0}^{\infty} e^{-2n}\). This can be recognized as a geometric series where each term is obtained by multiplying the previous term by a constant factor, \(r\). Here, the constant is \(e^{-2}\).

Step 2: Check for Convergence

To determine convergence, evaluate the absolute value of the common ratio, \(r = e^{-2}\). For a geometric series to converge, the condition \(|r| < 1\) must be satisfied. Calculate \(e^{-2}\) which is approximately 0.135, clearly less than 1. Hence, the series converges.

Step 3: Use the Formula for the Sum of a Convergent Geometric Series

The sum \(S\) of a convergent geometric series \(\sum_{n=0}^{\infty} ar^n\) is given by the formula \(S = \frac{a}{1 - r}\). Here \(a = 1\) (the first term of the series) and \(r = e^{-2}\).

Step 4: Plug Values into the Sum Formula

Substitute \(a = 1\) and \(r = e^{-2}\) into the formula: \[ S = \frac{1}{1 - e^{-2}} \]Calculate this to find the sum of the series.

Step 5: Simplify the Sum Expression

Calculating, we have:\[ S = \frac{1}{1 - e^{-2}} = \frac{1}{1 - 0.135} = \frac{1}{0.865} \approx 1.156. \]Thus, the sum of the series is approximately 1.156.

Key Concepts

Series Convergence

In mathematics, a series is a sum of a sequence of numbers. When you’re dealing with an infinite series, there’s a special interest in whether this sum has a finite value. This is called convergence. A series that grows without bound, or doesn’t settle to a fixed number, is said to diverge. For a geometric series, like the one in our exercise, it will converge under certain conditions. Specifically:

• The series converges if the common ratio's absolute value is less than 1.
• If the common ratio is equal to or greater than 1, the series diverges.

In our example, the series \(\sum_{n=0}^{\infty} e^{-2n}\) meets this condition because the common ratio, \(e^{-2}\), is approximately 0.135, which is less than 1. This tells us that the series converges and has a finite sum.

Common Ratio

The term "common ratio" is used in the context of geometric series. It is the fixed number that each term in the series is multiplied by to get the next term. For instance, in the exercise, the series \(\sum_{n=0}^{\infty} e^{-2n}\) has a common ratio of \(r = e^{-2}\). Identifying the common ratio is crucial for determining the behavior of the series.

Consider these points about common ratios:

• If \(|r| < 1\), the series will converge.
• If \(|r| \geq 1\), the series will diverge because the terms will not approach zero.

For our series, \(e^{-2}\) is a very small number, approximately 0.135, thus ensuring that the series terms decrease and lead to convergence.

Infinite Series

An infinite series is a sum of infinitely many terms. Unlike finite series, an infinite series does not have an easily measurable end, thus making it a bit more complex to analyze.

Here's what you need to know:

• The notation \(\sum_{n=0}^{\infty} a_n\) signifies an infinite series starting from \(n=0\) and extending forever.
• The crucial question is whether the sum approaches a finite number (convergence).
• Each additional term in a converging infinite series will get progressively smaller in absolute value.

In the exercise, the infinite series \(\sum_{n=0}^{\infty} e^{-2n}\) converges, meaning its sum can be calculated despite its infinite number of terms.

Sum of Series

The sum of a geometric series, especially when it converges, can be determined by a specific formula. For a convergent geometric series with a starting term \(a\) and common ratio \(r\), the sum \(S\) is given by \[ S = \frac{a}{1-r} \].

For instance, in our exercise:

• The series \(\sum_{n=0}^{\infty} e^{-2n}\) starts with \(a = 1\).
• The common ratio is \(r = e^{-2}\).

By substituting into the formula, you find the sum \(S = \frac{1}{1 - e^{-2}}\). Calculating this yields approximately 1.156. Thus, the entire series, from first to infinite term, sums to about 1.156.