Problem 38 - Calculus: Early Transcendentals

Question

Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$ \displaystyle \sum_{k = 0}^{\infty} (\sqrt 2)^{-k} $$

Step-by-Step Solution

Answer

The series is convergent and its sum is \( \frac{\sqrt{2}}{\sqrt{2} - 1} \).

Step 1: Identify the Series

The given series is \( \sum_{k=0}^{\infty} (\sqrt{2})^{-k} \). This can be rewritten as \( \sum_{k=0}^{\infty} \left( \frac{1}{\sqrt{2}} \right)^k \). This is a geometric series with the first term \( a = 1 \) and common ratio \( r = \frac{1}{\sqrt{2}} \).

Step 2: Check Convergence Conditions for Geometric Series

A geometric series \( \sum_{k=0}^{\infty} ar^k \) converges if the absolute value of the common ratio \( |r| < 1 \). In our series, \( |r| = \left| \frac{1}{\sqrt{2}} \right| = \frac{1}{\sqrt{2}} \), which is less than 1.

Step 3: Calculate the Sum Using the Geometric Series Formula

Since the series is convergent, we can calculate its sum using the formula for the sum of an infinite geometric series: \( S = \frac{a}{1-r} \). Substitute \( a = 1 \) and \( r = \frac{1}{\sqrt{2}} \) into the formula: \[ S = \frac{1}{1 - \frac{1}{\sqrt{2}}} \].

Step 4: Simplify the Expression for the Sum

First, find a common denominator to simplify the expression: \[ S = \frac{1}{\frac{\sqrt{2} - 1}{\sqrt{2}}} \]. Simplify by multiplying both the numerator and the denominator by \( \sqrt{2} \) to remove the fraction:\[ S = \frac{\sqrt{2}}{\sqrt{2} - 1} \].

Step 5: Further Simplification (Optional)

This result, \( \frac{\sqrt{2}}{\sqrt{2} - 1} \), can be left as is or further simplified by rationalizing the denominator, if needed, for clarity in some contexts.

Key Concepts

Infinite Series

An infinite series is a sum of infinitely many successive terms. We represent these series using the sigma notation \( \sum_{k=0}^{\infty} a_k \), where \( a_k \) denotes the terms of the series that depend on the index \( k \). Infinite series are fascinating because although they consist of numerous terms, they often result in a simple finite value when the series converges. In mathematics, infinite series frequently appear in calculus, physics, and engineering, aiding in calculations and modeling complex opportunities.

When dealing with infinite series, a key aspect to understand is whether the series converges, meaning that as you add up more and more terms, their sum approaches a specific value, or if it diverges, implying the sum does not settle on any particular number. Understanding convergence is crucial in determining the value of an infinite series.

Series Convergence

Series convergence is all about deciding if a series settles down, or converges, to a definite sum. To check convergence, especially for geometric series, we rely on the absolute value of the common ratio \(|r|\). If \(|r| < 1\), the series is convergent, meaning that as you keep adding terms together, you get closer to a certain number.

For instance, in the series \(\sum_{k=0}^{\infty} \left( \frac{1}{\sqrt{2}} \right)^k\), the common ratio \(r = \frac{1}{\sqrt{2}}\) is clearly smaller than 1. This guarantees convergence.

• If \(|r| > 1\), the terms keep getting larger, leading the series to diverge.
• If \(|r| = 1\), special tests are required since the standard convergence test is inconclusive.

Understanding if a series converges is essential to learn its behavior and potential applications. Mathematicians and scientists depend on this knowledge to find approximate results for otherwise complex or impossible calculations.

Geometric Series Formula

The geometric series formula is a powerful tool for calculating the sum of a convergent geometric series. When a geometric series converges, its sum \(S\) can be determined with the simple formula: \[S = \frac{a}{1-r}\] where \(a\) is the first term of the series and \(r\) is the common ratio. This formula applies when the absolute value of the common ratio \(|r|\) is less than 1.
In the exercise provided, the series \( \sum_{k=0}^{\infty} \left( \frac{1}{\sqrt{2}} \right)^k \) simplifies to have a first term \(a = 1\) and a common ratio \(r = \frac{1}{\sqrt{2}}\). Substituting these into the formula gives \[S = \frac{1}{1 - \frac{1}{\sqrt{2}}}\].
Let's break down this calculation:

• First, ensure convergence: \( \left| \frac{1}{\sqrt{2}} \right| < 1\).
• Then, substitute values into the formula.
• Lastly, simplify the expression to find the sum.

This process affords you a practical method for deducing the sum of terms in a geometric series, granting insight into the series' end behavior and offering crucial solutions to real-world applications.