Problem 49 - Thomas Calculus

Question

Which series in Exercises \(49-68\) converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. $$ \sum_{n=0}^{\infty}\left(\frac{1}{\sqrt{2}}\right)^{n} $$

Step-by-Step Solution

Answer

The series converges, and its sum is \( 2 + \sqrt{2} \).

Step 1: Identify the Series Type

The series given is \( \sum_{n=0}^{\infty} \left(\frac{1}{\sqrt{2}}\right)^n \). This is a geometric series because each term is derived by multiplying the previous term by a constant factor \( \frac{1}{\sqrt{2}} \). The general form of a geometric series is \( \sum_{n=0}^{\infty} ar^n \). Here, \( a = 1 \) (the first term) and \( r = \frac{1}{\sqrt{2}} \).

Step 2: Determine Convergence or Divergence

For a geometric series \( \sum_{n=0}^{\infty} ar^n \), the series converges if and only if the absolute value of the common ratio \( |r| < 1 \). Here, \( |r| = \left|\frac{1}{\sqrt{2}}\right| = \frac{1}{\sqrt{2}} \), which is less than 1. Therefore, the series converges.

Step 3: Find the Sum of the Convergent Series

The sum \( S \) of a convergent geometric series \( \sum_{n=0}^{\infty} ar^n \) is given by the formula \( S = \frac{a}{1-r} \). Substituting the values \( a = 1 \) and \( r = \frac{1}{\sqrt{2}} \), we have:\[ S = \frac{1}{1 - \frac{1}{\sqrt{2}}} = \frac{1}{\frac{\sqrt{2}-1}{\sqrt{2}}} = \frac{\sqrt{2}}{\sqrt{2} - 1}. \]

Step 4: Simplify the Sum Expression

The expression \( \frac{\sqrt{2}}{\sqrt{2} - 1} \) can be rationalized further by multiplying the numerator and denominator by the conjugate \( \sqrt{2} + 1 \):\[ S = \frac{\sqrt{2}(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{2 + \sqrt{2}}{1} = 2 + \sqrt{2}. \]Thus, the sum of the series is \( 2 + \sqrt{2} \).

Key Concepts

Series Convergence

A geometric series is a special kind of series where each term is multiplied by a constant called the common ratio. Determining if a series converges means finding out if the sum of its terms approaches a specific number as you continue to add more terms. For geometric series, convergence depends on the common ratio.

The key criterion is that the absolute value of the common ratio, denoted as \(|r|\), must be less than 1. If \(|r| \geq 1\), the series diverges, meaning it doesn't approach a particular value.

• If \(|r| < 1\), the series converges.
• If \(|r| \geq 1\), the series diverges.

In our example, the series \( \sum_{n=0}^{\infty} \left(\frac{1}{\sqrt{2}}\right)^n \) converges because \( \left| \frac{1}{\sqrt{2}} \right| \) is less than one, confirming it will sum to a finite number.

Sum of Infinite Series

Once we know a geometric series converges, we can find its sum using a simple formula. This formula helps calculate the total of all infinite terms of the series up to its convergence point.

The formula to find the sum \( S \) of a convergent geometric series is:\[ S = \frac{a}{1 - r} \]where \( a \) is the first term, and \( r \) is the common ratio.

For our specific series, the first term \( a \) is 1, and the common ratio \( r \) is \( \frac{1}{\sqrt{2}} \). Substituting those into our formula, we have:\[ S = \frac{1}{1 - \frac{1}{\sqrt{2}}} \]This expression simplifies to \( 2 + \sqrt{2} \), which is the sum of the series. Simplifying such formulas might involve rationalizing denominators, as seen here, by multiplying by the conjugate when necessary.

Common Ratio

The common ratio in a geometric series is a pivotal concept because it determines the behavior of the series as a whole. This value, denoted by \( r \), shows how each term relates to the previous one and dictates the series' convergence or divergence.

In practical terms, you find the common ratio by dividing any term by its predecessor. In the series \( \sum_{n=0}^{\infty} \left(\frac{1}{\sqrt{2}}\right)^n \), it is obvious as \( \frac{1}{\sqrt{2}} \).

• It's less than 1, which indicates that each subsequent term will be a fraction of the last.
• This shrinking pattern signifies convergence, leading the series to sum to a finite number.

Understanding the common ratio is crucial, as it governs whether our series behaves predictably or not, ensuring we can find its total sum if it converges.