Chapter 10: Problem 49
Which series converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. $$\sum_{n=0}^{\infty}\left(\frac{1}{\sqrt{2}}\right)^{n}$$
Short Answer
Expert verified
The series converges and its sum is \(2+\sqrt{2}\).
Step by step solution
01
Identify the Series Type
The series given is \(\sum_{n=0}^{\infty} a^n\) where \(a = \frac{1}{\sqrt{2}}\). This is a geometric series since the terms take the form \(a^n\).
02
Condition for Convergence
A geometric series \(\sum_{n=0}^{\infty} ar^n\) converges if the common ratio \(r\) satisfies \(|r| < 1\). Here, \(r = \frac{1}{\sqrt{2}}\), which is less than 1. Thus, the series converges.
03
Find the Sum of the Series
The sum of an infinite geometric series \(\sum_{n=0}^{\infty} ar^n\) is given by \(\frac{a}{1-r}\), where \(a = 1\) (as it is \(a^0 = 1\)) and \(r = \frac{1}{\sqrt{2}}\). Thus, the sum is \(\frac{1}{1-\frac{1}{\sqrt{2}}}\).
04
Simplify the Expression
Calculate \(1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}-1}{\sqrt{2}}\). Substitute into the sum formula: \(\frac{1}{\frac{\sqrt{2}-1}{\sqrt{2}}} = \frac{\sqrt{2}}{\sqrt{2}-1}\).
05
Rationalize the Denominator
Multiply the numerator and denominator by the conjugate of the denominator: \(\frac{\sqrt{2}}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{\sqrt{2}(\sqrt{2}+1)}{2-1} = 2+\sqrt{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Geometric Series
A geometric series is a sequence of terms where each term after the first is found by multiplying the previous term by a fixed, non-zero number known as the common ratio. This type of series can be written in a general form
If \( |r| < 1 \), the series converges, and if \( |r| \geq 1 \), the series diverges.
In our example, the series \( \sum_{n=0}^{\infty} \left(\frac{1}{\sqrt{2}}\right)^{n} \) is geometric with a common ratio of \( \frac{1}{\sqrt{2}} \). Since \( \left| \frac{1}{\sqrt{2}} \right| < 1 \), the series converges.
- \( a + ar + ar^2 + ar^3 + \dots \,\)
If \( |r| < 1 \), the series converges, and if \( |r| \geq 1 \), the series diverges.
In our example, the series \( \sum_{n=0}^{\infty} \left(\frac{1}{\sqrt{2}}\right)^{n} \) is geometric with a common ratio of \( \frac{1}{\sqrt{2}} \). Since \( \left| \frac{1}{\sqrt{2}} \right| < 1 \), the series converges.
Infinite Series
An infinite series involves the sum of infinitely many terms. It's defined not merely as a sum but as a limit of a sequence of partial sums. Understanding infinite series helps in exploring important concepts in calculus and analysis.
An infinite series is often written as \( \sum_{n=0}^{\infty} a_n \), where \( a_n \) represents the sequence terms. To determine whether an infinite series converges, we observe whether the sequence of partial sums has a finite limit. For geometric series, particularly, the rule \( |r| < 1 \) indicates convergence.
Our exercise deals with such an infinite series where terms are determined by a geometric progression. Given its common ratio of \( \frac{1}{\sqrt{2}} \), the series converges because the ratio is effectively captured within the range that ensures convergence.
An infinite series is often written as \( \sum_{n=0}^{\infty} a_n \), where \( a_n \) represents the sequence terms. To determine whether an infinite series converges, we observe whether the sequence of partial sums has a finite limit. For geometric series, particularly, the rule \( |r| < 1 \) indicates convergence.
Our exercise deals with such an infinite series where terms are determined by a geometric progression. Given its common ratio of \( \frac{1}{\sqrt{2}} \), the series converges because the ratio is effectively captured within the range that ensures convergence.
Sum of Series
The sum of a convergent series can be an intriguing topic, especially when evaluating infinite series. For a geometric series that meets the convergence condition \( |r| < 1 \), the sum is calculated using the formula
This formula promptly gives us the sum \( \frac{1}{1 - \frac{1}{\sqrt{2}}} \). After simplifying, the calculation steps yield a sum of \( 2 + \sqrt{2} \).
This result confirms that despite dealing with infinitely many terms, when adhering to conditions of convergence, even infinite series can be nicely boiled down to a finite sum!
- \( \frac{a}{1-r} \)
This formula promptly gives us the sum \( \frac{1}{1 - \frac{1}{\sqrt{2}}} \). After simplifying, the calculation steps yield a sum of \( 2 + \sqrt{2} \).
This result confirms that despite dealing with infinitely many terms, when adhering to conditions of convergence, even infinite series can be nicely boiled down to a finite sum!