Problem 14 - A First Course in Complex Analysis with Applications

Question

Use the sequence of partial sums to show that the given series is convergent. $$ \sum_{k=1}^{\infty} \frac{i}{k(k+1)} $$

Step-by-Step Solution

Answer

The series converges to i.

Step 1: Identify the General Term of the Series

The general term of the given series is \( a_k = \frac{i}{k(k+1)} \). The aim is to express this term in a form that allows us to find the sequence of partial sums.

Step 2: Perform Partial Fraction Decomposition

The term \( \frac{i}{k(k+1)} \) can be decomposed using partial fractions. We express it as:\[\frac{i}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}\]Multiplying through by \( k(k+1) \) and equating coefficients yields:\[i = A(k+1) + Bk\]\[i = Ak + A + Bk \Rightarrow Ak + Bk + A = i\]Comparing coefficients of like terms, we have:\[A + B = 0 \A = i \]Solving these gives:\( A = i \), \( B = -i \), so\[\frac{i}{k(k+1)} = \frac{i}{k} - \frac{i}{k+1}\]

Step 3: Write the Partial Sum

The nth partial sum \( S_n \) of the series is a sum of the first n terms:\[S_n = \left( \frac{i}{1} - \frac{i}{2} \right) + \left( \frac{i}{2} - \frac{i}{3} \right) + \cdots + \left( \frac{i}{n} - \frac{i}{n+1} \right)\]

Step 4: Simplify the Partial Sum Using Telescoping

Notice that most terms will cancel each other in the series, leaving us with:\[S_n = i \left( 1 - \frac{1}{n+1} \right)\]This is because all intermediate terms from \( \frac{i}{2} \) to \( \frac{i}{n} \) cancel out.

Step 5: Determine the Limit of the Partial Sum as n Approaches Infinity

To determine whether the series converges, find the limit of \( S_n \) as \( n \to \infty \):\[\lim_{n \to \infty} S_n = \lim_{n \to \infty} i \left( 1 - \frac{1}{n+1} \right) = i\]Since this limit is finite, the series converges.

Key Concepts

Partial Sums

In the context of series, a partial sum is the sum of the first n terms of a sequence. When you consider partial sums, your goal is to evaluate the behavior of a series as you add more and more terms. Essentially, it's like snapping pieces together to build a larger picture.

• To find the nth partial sum, you add up the terms from the first term up to the nth term.
• This is denoted as \( S_n \), where \( n \) is the number of terms being summed.

In the original exercise, the sequence of partial sums helps us assess if the series converges as the number of terms, \( n \), grows infinitely. By examining these partial sums, we notice the cumulative effect of the terms and how they approach a particular value. This process is pivotal in determining the overall behavior of the series as the indices increase.

Partial Fraction Decomposition

Partial fraction decomposition is a handy technique in algebra that simplifies complex rational expressions into simpler fractions. This approach is especially useful when you want to break down a term that otherwise seems difficult to handle or analyze.

• In the given series, the term \( \frac{i}{k(k+1)} \) is decomposed into a sum of simpler fractions.
• This method involves expressing the term as the sum of two separate fractions: \( \frac{A}{k} + \frac{B}{k+1} \).

Through this process, you determine the values of \( A \) and \( B \) by comparing terms and solving simple equations, resulting in \( A = i \) and \( B = -i \). This decomposition is crucial as it prepares the series for the telescoping process, which simplifies finding the sum of the series.

Telescoping Series

A telescoping series is a special type of series where many terms cancel each other out, making it easier to find the sum. This technique resembles the compact nature of a telescope, where sections slide into each other, reducing total length.

• When you apply telescoping to a series, it reveals how terms within the series can cancel out sequentially.
• In this exercise, after decomposition, most terms cancel out except for the very first and the last term of the sum.

By simplifying the series with the telescoping method, you end up with a much simpler expression for the partial sum: \( S_n = i \left( 1 - \frac{1}{n+1} \right) \). This simplification facilitates the next steps in analyzing the convergence of the series, emphasizing the elegance and efficiency of telescoping.

Limit of Sequence

Finding the limit of a sequence helps us understand the behavior of a series as the number of terms goes to infinity. It's the process of answering, "What happens when we keep adding more and more terms?"

• The limit evaluates what the sequence looks like as \( n \) becomes very large.
• In the solution, the limit is calculated from the expression \( S_n = i \left( 1 - \frac{1}{n+1} \right) \).

Here, as \( n \to \infty \), \( \frac{1}{n+1} \to 0 \), ensuring that the partial sum approaches \( i \). This limit shows that the series converges, as the sum of the infinitely many terms doesn't explode to infinity or oscillate wildly, but instead settles at a fixed value, \( i \). This is a fundamental step in proving the convergence of infinite series.