/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Use the sequence of partial sums... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 14

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

Short Answer

Expert verified
The series converges to i.

Step by step solution

01

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.
02

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\[\frac{i}{k(k+1)} = \frac{i}{k} - \frac{i}{k+1}\]
03

Write the Partial Sum

The nth partial sum \( S_n \) of the series is a sum of the first n terms:\[S_n = \left( \frac{i}{1} - \frac{i}{2} \right) + \left( \frac{i}{2} - \frac{i}{3} \right) + \cdots + \left( \frac{i}{n} - \frac{i}{n+1} \right)\]
04

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.
05

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine the zeros and their order for the given function. $$ f(z)=z^{4}+z^{2} $$

Use an appropriate Laurent series to find the indicated residue. $$ f(z)=e^{-2 / z^{2}} ; \operatorname{Res}(f(z), 0) $$

Use an appropriate Laurent series to find the indicated residue. $$ f(z)=\frac{1}{z^{3}(1-z)^{3}} ; \operatorname{Res}(f(z), 0) $$

Evaluate the given trigonometric integral. $$ \int_{0}^{\pi} \frac{1}{2-\cos \theta} d \theta[\text { Hint }: \text { Let } t=2 \pi-\theta .] $$

Discuss how to evaluate \(\oint_{C} e^{1 / z} \sin \left(\frac{1}{z}\right) d z\), where \(C\) is \(|z|=1\). Carry out your ideas.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.