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Chapter 9: Problem 7

Determine whether the series converges, and if so find its sum. $$ \sum_{k=1}^{\infty} \frac{1}{(k+2)(k+3)} $$

Short Answer

Expert verified
The series converges to \( \frac{1}{3} \).

Step by step solution

01

Decompose the fraction

Notice that the term in the series can be decomposed using partial fractions. Let's write \[ \frac{1}{(k+2)(k+3)} = \frac{A}{k+2} + \frac{B}{k+3}. \]By clearing the fractions we get\[ 1 = A(k+3) + B(k+2). \]Expanding and rearranging terms gives: \[ 1 = (A+B)k + (3A + 2B). \]This must hold for all \(k\), so set the coefficients equal:\[ A + B = 0 \]and\[ 3A + 2B = 1. \]
02

Solve for the coefficients

From \(A + B = 0\), we have \(B = -A\). Substitute \(B = -A\) into \(3A + 2B = 1\):\[ 3A + 2(-A) = 1 \]\[ 3A - 2A = 1 \]\[ A = 1. \]Now, substitute \(A = 1\) into \(B = -A\):\[ B = -1. \]Thus, \(\frac{1}{(k+2)(k+3)} = \frac{1}{k+2} - \frac{1}{k+3}.\)
03

Recognize the telescoping nature

The series terms are now\[ \sum_{k=1}^{\infty} \left( \frac{1}{k+2} - \frac{1}{k+3} \right). \]This is a telescoping series, which means when expanded, most terms cancel with each other. Let's write out the first few terms:\[ \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{6} \right) + \dots \]
04

Determine the convergence and find the sum

In a telescoping series, cancellation occurs such that only the first few and the last few terms remain:- The \( \frac{1}{3} \) term from the beginning does not get canceled.- As \(k\) tends to infinity, the \( \frac{1}{k+3} \) approaches zero.Therefore, the sum of the series is the first term that did not get canceled, which is \( \frac{1}{3} \). Therefore, the series converges and its sum is \( \frac{1}{3} \).

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Key Concepts

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

Partial Fractions
Partial fractions is a technique used to break down complex fractions into simpler individual fractions to make problems easier to handle, especially in calculus and algebra.When you have a fraction with a product in the denominator, as in \( \frac{1}{(k+2)(k+3)} \), it can be expressed as a sum of two separate fractions, \( \frac{A}{k+2} + \frac{B}{k+3} \).Here鈥檚 a simple breakdown of how it works:
  • Start with expressing the original fraction as two partial fractions.
  • Clear the denominators by multiplying through by \( (k+2)(k+3) \).
  • This should give you an equation in terms of \( k \).
  • By matching coefficients for each \( k \) term, you set up equations to solve for unknowns \( A \) and \( B \).
This method helps break down the fraction into a format that's easier to work with when summing a series, and it's a crucial step for recognizing telescoping series.
Series Convergence
Series convergence refers to whether the sum of an infinite series approaches a specific value. When a series converges, it means that as more terms are added, the total sum gets closer to a particular number.To determine convergence, we often analyze the general term: in this problem, \( \frac{1}{(k+2)(k+3)} \).Key points include:
  • A series converges if the sequence of its partial sums approaches a limit as the number of terms increases.
  • In telescoping series, most terms cancel out, simplifying determining convergence.
  • Tests like the Ratio Test or Comparison Test can be used, but special series like telescoping often reveal convergence through simplification.
For the given series, we identified it as telescoping, where many terms cancel, letting us see that only certain terms in the beginning impact the sum as more terms are added.
Infinite Series
An infinite series is a sum of infinitely many terms. It has a start but no end, and understanding whether it converges or diverges is key to working with it effectively.Some aspects to remember:
  • Each term in an infinite series is a function of an index which runs to infinity, such as \( k \) in \( \frac{1}{(k+2)(k+3)} \).
  • Just because a series is infinite doesn't mean it lacks a finite sum鈥攃onvergent series have finite sums.
  • Different methods and series types (like geometric, harmonic, etc.) require different techniques to evaluate the sum.
For example, a telescoping series like the one in this exercise is a specific kind of infinite series where terms cancel successively. This simplifies evaluation and helps easily determine if it converges and to what sum. Understanding how to manage and sum infinite series is a powerful tool in mathematics.

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Most popular questions from this chapter

Suppose that the power series \(\sum c_{k}\left(x-x_{0}\right)^{k}\) has a finite radius of convergence \(R,\) and the power series \(\sum_{k} d_{k}\left(x-x_{0}\right)^{k}\) has a radius of convergence of \(+\infty\) What can you say about the radius of convergence of \(\sum\left(c_{k}+d_{k}\right)\left(x-x_{0}\right)^{k}\) ? Explain your reasoning.

Use a CAS to confirm that $$ \sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} \quad \text { and } \quad \sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90} $$ and then use these results in each part to find the sum of the series. $$ \text { (a) } \sum_{k=1}^{\infty} \frac{3 k^{2}-1}{k^{4}} \quad \text { (b) } \sum_{k=3}^{\infty} \frac{1}{k^{2}} \quad \text { (c) } \sum_{k=2}^{\infty} \frac{1}{(k-1)^{4}} $$

Apply the divergence test and state what it tells you about the series. $$ \begin{array}{ll}{\text { (a) } \sum_{k=1}^{\infty} \frac{k}{e^{k}}} & {\text { (b) } \sum_{k=1}^{\infty} \ln k} \\ {\text { (c) } \sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}} & {\text { (d) } \sum_{k=1}^{\infty} \frac{\sqrt{k}}{\sqrt{k}+3}}\end{array} $$

Find the radius of convergence and the interval of convergence. $$ \sum_{k=0}^{\infty} \frac{x^{k}}{1+k^{2}} $$

Find the radius of convergence and the interval of convergence. $$ \sum_{k=2}^{\infty}(-1)^{k+1} \frac{x^{k}}{k(\ln k)^{2}} $$
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