Problem 141
Question
The sum of the series \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\ldots \ldots \ldots\) upto $$ \infty \text { is equal to } $$ [2003] (A) \(2 \log _{e} 2\) (B) \(\log _{2} 2-1\) (C) \(\log _{e} 2\) (D) \(\log _{e}\left(\frac{4}{e}\right)\)
Step-by-Step Solution
Verified Answer
The sum of the series is \( \log_e 2 \) (option C).
1Step 1: Recognize the Pattern
The given series is \( \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} - \ldots \). This series is an alternating series where each term can be expressed as \( (-1)^{n+1} \frac{1}{n(n+1)} \). Noticing this alternating pattern is crucial for further evaluation.
2Step 2: Decompose the General Term
Notice \( \frac{1}{n(n+1)} \) can be decomposed using partial fractions. We write \( \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1} \). Solving for \(A\) and \(B\), we find that \(A = 1\) and \(B = -1\). So, \( \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \).
3Step 3: Express the Series with Partial Fractions
Substitute \(\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\) into the series. The series becomes: \[ \left( \frac{1}{1} - \frac{1}{2} \right) - \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) - \ldots \] aimplying that it telescopes neatly.
4Step 4: Understand the Telescoping Nature
In a telescoping series, most terms cancel out with one another. Expanding the series, we see that each positive term cancels with the negative term of the subsequent fraction. The only terms that do not cancel, therefore, belong to the start and the end of the series.
5Step 5: Evaluate the Limit of the Telescoping Series
Given this is an infinite series, as \( n \) approaches infinity, the series reduces to: \[ 1 + \lim_{n \to \infty} \left(-\frac{1}{n+1}\right) = 1 - 0 = 1 \] due to the negative fraction decreasing towards zero.
6Step 6: Find Equivalent Expression for the Sum
The value of the infinite series is 1. This should match with one of the given options. Recognize that the natural logarithm expressions can express this value: \( \log_e 2 \) is approximately 0.693, thus the final correct option is \( \log_e 2 \).
Key Concepts
Alternating SeriesTelescoping SeriesPartial Fractions Decomposition
Alternating Series
An **alternating series** is a series in which the signs of the terms alternate. In the exercise at hand, the series consists of terms like \( \frac{1}{1 \cdot 2} \), \( -\frac{1}{2 \cdot 3} \), and so forth, where even-indexed terms are negative. This pattern is often expressed with \( (-1)^{n+1} \), indicating that terms switch between positive and negative as \( n \) increases.
Alternating series can be tricky but are important in mathematics as they often converge even when an absolute series does not. The Alternating Series Test can help determine convergence: if the terms decrease in absolute value and approach zero, the alternating series converges. This property is crucial in many mathematical and physical applications, providing insight into system stability in fields such as engineering and physics.
Alternating series can be tricky but are important in mathematics as they often converge even when an absolute series does not. The Alternating Series Test can help determine convergence: if the terms decrease in absolute value and approach zero, the alternating series converges. This property is crucial in many mathematical and physical applications, providing insight into system stability in fields such as engineering and physics.
Telescoping Series
A **telescoping series** is one where successive terms cancel each other out. This cancellation effect is evident in the given series when expressed with partial fractions. The terms \( \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \) lead to a chain of positive and negative pairs that neatly cancel each other.
The beauty of telescoping series lies in their simplicity of evaluation. After the middle terms cancel, only the initial and final terms matter, often making such series straightforward to solve. For infinite telescoping series, the remaining expressions at the extremes can often lead directly to the series' sum, as it simplifies the calculation by limiting the terms that contribute to the final result.
The beauty of telescoping series lies in their simplicity of evaluation. After the middle terms cancel, only the initial and final terms matter, often making such series straightforward to solve. For infinite telescoping series, the remaining expressions at the extremes can often lead directly to the series' sum, as it simplifies the calculation by limiting the terms that contribute to the final result.
Partial Fractions Decomposition
**Partial fractions decomposition** is a technique used to break down complex fractions into simpler, more easily managed components. It's especially useful in calculus for integrating rational functions. In the series, the term \( \frac{1}{n(n+1)} \) is broken down into \( \frac{1}{n} - \frac{1}{n+1} \), a form that is much easier to work with.
To achieve this decomposition, you solve for constants that allow you to express the original fraction as a sum or difference of simpler fractions. For \( \frac{1}{n(n+1)} \), identifying \( A \) and \( B \) such that \( \frac{A}{n} + \frac{B}{n+1} = \frac{1}{n(n+1)} \), yields \( A = 1 \) and \( B = -1 \). This process simplifies the handling of the fraction and, in the context of series, prepares it for telescoping, ultimately aiding in the summation process.
To achieve this decomposition, you solve for constants that allow you to express the original fraction as a sum or difference of simpler fractions. For \( \frac{1}{n(n+1)} \), identifying \( A \) and \( B \) such that \( \frac{A}{n} + \frac{B}{n+1} = \frac{1}{n(n+1)} \), yields \( A = 1 \) and \( B = -1 \). This process simplifies the handling of the fraction and, in the context of series, prepares it for telescoping, ultimately aiding in the summation process.
Other exercises in this chapter
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