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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
  • \( a + ar + ar^2 + ar^3 + \dots \,\)
where \( a \) is the first term and \( r \) is the common ratio. When we talk about an infinite geometric series, it means that there are an infinite number of terms. The convergence or divergence of a geometric series mainly depends on the value of \( r \).
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.
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
  • \( \frac{a}{1-r} \)
In this formula, \( a \) is the initial term, and \( r \) is the common ratio. In our exercise, \( a = 1 \) for \( a^0 = 1 \), and \( r = \frac{1}{\sqrt{2}} \).
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!

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Most popular questions from this chapter

According to the Alternating Series Estimation Theorem, how many terms of the Taylor series for \(\tan ^{-1} 1\) would you have to add to be sure of finding \(\pi / 4\) with an error of magnitude less than \(10^{-3} ?\) Give reasons for your answer.

Use the definition of \(e^{i \theta}\) to show that for any real numbers \(\theta, \theta_{1}\) and \(\theta_{2}\) a. \(\quad e^{i \theta_{1}} e^{i \theta_{2}}=e^{i\left(\theta_{1}+\theta_{2}\right)}\) b. \(e^{-i \theta}=1 / e^{i \theta}\)

Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals. Step 1: Plot the function over the specified interval. Step 2: Find the Taylor polynomials \(P_{1}(x), P_{2}(x),\) and \(P_{3}(x)\) at \(\bar{x}=0\) Step 3: Calculate the ( \(n+1\) )st derivative \(f^{(n+1)}(c)\) associated with the remainder term for each Taylor polynomial. Plot the derivative as a function of \(c\) over the specified interval and estimate its maximum absolute value, \(M\) Step 4: Calculate the remainder \(R_{n}(x)\) for each polynomial. Using the estimate \(M\) from Step 3 in place of \(f^{(n+1)}(c),\) plot \(R_{n}(x)\) over the specified interval. Then estimate the values of \(x\) that answer question (a). Step 5: Compare your estimated error with the actual error \(E_{n}(x)=\left|f(x)-P_{n}(x)\right|\) by plotting \(E_{n}(x)\) over the specified interval. This will help answer question (b). Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5. $$f(x)=(\cos x)(\sin 2 x), \quad|x| \leq 2$$

Integrate the binomial series for \(\left(1-x^{2}\right)^{-1 / 2}\) to show that for \(|x|<1\) $$ \sin ^{-1} x=x+\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot(2 n)} \frac{x^{2 n+1}}{2 n+1} $$

Replace \(x\) by \(-x\) in the Taylor series for \(\ln (1+x)\) to obtain a series for \(\ln (1-x)\). Then subtract this from the Taylor series for \(\ln (1+x)\) to show that for \(|x|<1\) $$\ln \frac{1+x}{1-x}=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right)$$

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