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Magazine Chapter 19: Problem 14
Use the sequence of partial sums to show that the given series is convergent. \(\sum_{k=2}^{\infty} \frac{i}{k(k+1)}\)
Short Answer
Expert verified
The series converges to \( \frac{i}{2} \).
Step by step solution
01
Understanding the Series
We are given the series \( \sum_{k=2}^{\infty} \frac{i}{k(k+1)} \). To determine if this series is convergent, we need to find the sequence of partial sums \( S_n = \sum_{k=2}^{n} \frac{i}{k(k+1)} \).
02
Express the General Term
Recognize that \( \frac{i}{k(k+1)} \) can be simplified using partial fraction decomposition. This means we express \( \frac{i}{k(k+1)} \) as \( \frac{A}{k} + \frac{B}{k+1} \). Solving gives \( A = i \) and \( B = -i \). Thus, \( \frac{i}{k(k+1)} = \frac{i}{k} - \frac{i}{k+1} \).
03
Formulate the Partial Sum
Using the decomposition from Step 2, write the partial sum \( S_n \) as: \[S_n = \sum_{k=2}^{n} \left( \frac{i}{k} - \frac{i}{k+1} \right).\]
04
Identify the Telescope
Notice that the series is telescoping. Most terms cancel each other: \[S_n = \left( \frac{i}{2} - \frac{i}{3} \right) + \left( \frac{i}{3} - \frac{i}{4} \right) + \ldots + \left( \frac{i}{n} - \frac{i}{n+1} \right).\] Only the first part \( \frac{i}{2} \) and the last part \( \frac{i}{n+1} \) do not get canceled out.
05
Simplify the Partial Sum
Simplify the expression for \( S_n \):\[S_n = \frac{i}{2} - \frac{i}{n+1}.\] This simplification arises because of the telescoping nature of the series.
06
Evaluate the Limit of the Partial Sum
To determine convergence, compute \( \lim_{n \to \infty} S_n \):\[\lim_{n \to \infty} \left( \frac{i}{2} - \frac{i}{n+1} \right) = \frac{i}{2} - 0 = \frac{i}{2}.\] The limit exists and is finite, so the series converges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Sums
When tackling series problems, partial sums are a useful tool in analysis. They are essentially the cumulative happening of a series up to a certain point. For the series \( \sum_{k=2}^{n} \frac{i}{k(k+1)} \), the partial sum \( S_n \) is calculated by adding the sequence elements up to an index \( n \).
This concept helps visibly track how a series behaves as more terms are considered.
This concept helps visibly track how a series behaves as more terms are considered.
- The partial sum \( S_2 \) for example, would include only the first term of this sequence.
- Progressing to higher \( n \) means including more terms in our sum, giving us \( S_n \).
Partial Fraction Decomposition
Partial fraction decomposition is a technique used to express complex fractions into simpler components, making calculations easier. The series element \( \frac{i}{k(k+1)} \) can be split into simpler fractions:
It involves finding constants such that the fraction can be expressed as \( \frac{A}{k} + \frac{B}{k+1} \).
It involves finding constants such that the fraction can be expressed as \( \frac{A}{k} + \frac{B}{k+1} \).
- Here, solving the equation for constants \( A \) and \( B \) gives \( A = i \) and \( B = -i \).
- This allows rewriting the term as \( \frac{i}{k} - \frac{i}{k+1} \), which simplifies the process of summation.
Telescoping Series
A telescoping series is one where many terms cancel each other out when the sequence is expanded. In our series, the decomposition \( \frac{i}{k} - \frac{i}{k+1} \) leads to such a simplification.
This means that:
This means that:
- Terms like \( \frac{i}{3} \) and \( -\frac{i}{3} \) cancel each other.
- Ultimately, this leaves a few terms like \( \frac{i}{2} \) and \( \frac{i}{n+1} \) which do not get cancelled.
Limit Evaluation
Checking the convergence of a series involves determining the limit of its partial sums as \( n \) becomes very large (approaches infinity). For the given series, we evaluate\[ \lim_{n \to \infty} \left( \frac{i}{2} - \frac{i}{n+1} \right). \]
This calculation leverages the telescoped form, where only a few terms remain unchecked. As \( n \rightarrow \infty \), the term \( \frac{i}{n+1} \) diminishes to zero. Hence,
This calculation leverages the telescoped form, where only a few terms remain unchecked. As \( n \rightarrow \infty \), the term \( \frac{i}{n+1} \) diminishes to zero. Hence,
- The limit becomes \( \frac{i}{2} - 0 = \frac{i}{2} \).
- The finite limit indicates that the series indeed converges, with the sum converging to \( \frac{i}{2} \).
