Problem 24
Question
Express each of the sums in closed form. Wherever possible, give a numerical approximation of the sum, rounded off to 3 decimal places. $$ \frac{1}{e}+\frac{2}{e^{2}}+\frac{4}{e^{3}}+\cdots+\frac{2^{n}}{e^{n+1}} $$
Step-by-Step Solution
Verified Answer
The closed form of the series sum is \(S = e^2/(e^2 - 2e)\). The numerical approximation of the sum, rounded to 3 decimal places, will be obtained by substituting the approximation value of ‘e’ as 2.718 in the above expression.
1Step 1: Identify the type of series
The given series is a geometric series because the ratio of successive terms is constant. The first term \(a = 1/e\), and the common ratio \(r = 2/e\).
2Step 2: Apply the formula for geometric series sum
For a geometric series, the formula for finding the sum is \(S = a/(1−r)\), where 'S' is the sum of the series, 'a' is the first term, and 'r' is the common ratio. Following this formula, the sum of the series is \(S = 1/(e(1 - 2/e))\).
3Step 3: Simplify the expression
Simplify the denominator using the basic properties of fractions, to obtain the sum in closed form. Simplifying gives, \(S = 1/((e^2 - 2e)/e^2) = e^2/(e^2 - 2e)\).
4Step 4: Calculate numerical approximation
After finding the closed form of the sum, use the approximation value of ‘e’ as 2.718 (up to 3 decimal places), to obtain a numerical approximation of the sum \(S\). This gives, \(S = (2.718^2)/(2.718^2 - 2*2.718)\).
5Step 5: Round off to 3 decimal places
After performing the numerical computation, round off the result to 3 decimal places to get the final answer.
Key Concepts
Closed Form ExpressionSeries SummationNumerical Approximation
Closed Form Expression
Understanding the concept of a closed form expression can greatly simplify your calculations with series. A closed form expression is a mathematical representation of a series that does not require an iterative process to find subsequent terms. It allows you to determine the sum of a series directly. This is particularly useful in mathematics because working with long series terms individually can be cumbersome. Instead, a closed form expression provides a succinct and efficient way to find the entire series sum without needing to add each term separately.
In the original exercise, the challenge was to express the sum of a geometric series in closed form. The series involves terms like \( \frac{1}{e}, \frac{2}{e^2}, \) and so on. To find a closed form expression, we use the formula for the sum of a geometric series \( S = \frac{a}{1-r} \), where \( a \) is the first term and \( r \) is the common ratio. By applying this formula, we derived the closed form as \( \frac{e^2}{e^2 - 2e} \). This expression allows us to compute the sum of the series effectively without adding each individual term.
In the original exercise, the challenge was to express the sum of a geometric series in closed form. The series involves terms like \( \frac{1}{e}, \frac{2}{e^2}, \) and so on. To find a closed form expression, we use the formula for the sum of a geometric series \( S = \frac{a}{1-r} \), where \( a \) is the first term and \( r \) is the common ratio. By applying this formula, we derived the closed form as \( \frac{e^2}{e^2 - 2e} \). This expression allows us to compute the sum of the series effectively without adding each individual term.
Series Summation
Series summation involves adding a sequence of numbers sequentially. This concept is widely used in mathematics, especially when dealing with infinite sequences or when evaluating patterns within series. Unlike simple addition, series summation often requires specific formulas or methods to accurately determine the total sum. This is key to understanding many mathematical phenomena, especially in calculus and higher math.
In a geometric series, like the one we have in the original exercise, each term differs by a constant ratio. Recognizing this pattern allows us to use the geometric series sum formula \( S = \frac{a}{1-r} \). This formula simplifies the summation process, providing a quick way to sum up all terms by knowing just the first term and the common ratio. When we applied it to our exercise, it confirmed the sum of terms like \( \frac{1}{e}, \frac{2}{e^2}, \frac{4}{e^3}, \ldots \) can be accurately calculated using a structured approach.
In a geometric series, like the one we have in the original exercise, each term differs by a constant ratio. Recognizing this pattern allows us to use the geometric series sum formula \( S = \frac{a}{1-r} \). This formula simplifies the summation process, providing a quick way to sum up all terms by knowing just the first term and the common ratio. When we applied it to our exercise, it confirmed the sum of terms like \( \frac{1}{e}, \frac{2}{e^2}, \frac{4}{e^3}, \ldots \) can be accurately calculated using a structured approach.
Numerical Approximation
Numerical approximation refers to estimating a numerical value when an exact solution is impractical to obtain or overly complex to express precisely. It often involves rounding off numbers to a convenient or necessary degree of accuracy. This technique is vital in many areas, including engineering, computer science, and physics, where exact values may not be necessary, and efficiency is paramount.
In the context of the given exercise, we derived a closed form for the series and then proceeded to find a numerical approximation. Given that \( e \approx 2.718 \), we substituted this approximate value to estimate the sum of the series as \( S \approx \frac{2.718^2}{2.718^2 - 2 \times 2.718} \). Finally, performing the calculations and rounding the result to three decimal places, we obtain a practical numerical approximation which simplifies and summarizes the result effectively.
In the context of the given exercise, we derived a closed form for the series and then proceeded to find a numerical approximation. Given that \( e \approx 2.718 \), we substituted this approximate value to estimate the sum of the series as \( S \approx \frac{2.718^2}{2.718^2 - 2 \times 2.718} \). Finally, performing the calculations and rounding the result to three decimal places, we obtain a practical numerical approximation which simplifies and summarizes the result effectively.
Other exercises in this chapter
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