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Chapter 9: Problem 60

For the following telescoping series, find a formula for the nth term of the sequence of partial sums \(\left\\{S_{n}\right\\} .\) Then evaluate lim \(S_{n}\) to obtain the value of the series or state that the series diverges. \(^{n \rightarrow \infty}\) $$\sum_{k=3}^{\infty} \frac{2}{(2 k-1)(2 k+1)}$$

Short Answer

Expert verified
Using a telescoping series method, we have found the formula for the nth term of the sequence of partial sums to be: $$S_n = \frac{1}{7}\left(\frac{1}{5} - \frac{1}{2n+1}\right)$$ Evaluating the limit as n approaches infinity, we found the value of the series to be: $$\lim_{n \rightarrow \infty} S_n = \frac{1}{35}$$

Step by step solution

01

Write down the series

For this exercise, we are given the series: $$\sum_{k=3}^{\infty} \frac{2}{(2 k-1)(2 k+1)}$$ Our goal is to find a formula for the nth term of the sequence of partial sums and then evaluate the limit as \(n\rightarrow \infty\).
02

Create a telescoping form for the series

The first step is to find the telescoping form for the series. The idea behind telescoping is to write each term as the difference between two simpler terms, and when summed, most terms should cancel each other out as \(n\rightarrow \infty\). We can apply partial fraction decomposition to rewrite the given series as: $$\frac{2}{(2 k-1)(2 k+1)} = \frac{A}{2k-1} + \frac{B}{2k+1}$$ Multiplying both sides by the denominator, we have: $$2 = A(2k+1) + B(2k-1)$$ Solving for A and B, we can try some values of k. For convenience, let's try k = 3: $$A(3*2 + 1) = 2 \Rightarrow A = \frac{1}{7}$$ Now let's try k = 2: $$B(2*2 - 1) = 2 \Rightarrow B = \frac{1}{3}$$ So, we have the telescoping form as: $$\frac{2}{(2 k-1)(2 k+1)} = \frac{1}{7}\left(\frac{1}{2k-1} - \frac{1}{2k+1}\right)$$
03

Find a formula for the nth term of the sequence of partial sums

Now that we have the telescoping form, let's find the formula for the nth term of the sequence of partial sums: $$S_n = \sum_{k=3}^{n} \frac{1}{7}\left(\frac{1}{2k-1} - \frac{1}{2k+1}\right)$$ $$S_n = \frac{1}{7}\left(\left(\frac{1}{5} - \frac{1}{7}\right) + \left(\frac{1}{7} - \frac{1}{9}\right) + \ldots + \left(\frac{1}{2n-1} - \frac{1}{2n+1}\right)\right)$$ The terms inside the parentheses should cancel each other out: $$S_n = \frac{1}{7}\left(\frac{1}{5} - \frac{1}{2n+1}\right)$$
04

Evaluate the limit as n approaches infinity

Finally, we can evaluate the limit of the series as \(n \rightarrow \infty\): $$\lim_{n \rightarrow \infty} S_n = \lim_{n \rightarrow \infty} \frac{1}{7}\left(\frac{1}{5} - \frac{1}{2n+1}\right)$$ Since \(\frac{1}{2n+1}\) approaches 0 as \(n \rightarrow \infty\), we have: $$\lim_{n \rightarrow \infty} S_n = \frac{1}{7}\left(\frac{1}{5}\right) = \frac{1}{35}$$ Thus, the value of the telescoping series is \(\frac{1}{35}\).

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Key Concepts

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

Partial sums
In the context of series, partial sums refer to the sum of the first 'n' terms of a sequence. It's a way to approximate the total sum of an infinite series by summing a finite portion of it. Consider a series represented as \(\sum_{k=3}^{\infty} a_k\).The partial sum is denoted as \(S_n\), where \(S_n = \sum_{k=3}^{n} a_k\). By looking at the partial sums sequence, \(\{ S_1, S_2, S_3, \ldots, S_n \}\), we can analyze the behavior of a series as more terms are added.
  • A key feature of partial sums is they enable us to observe if adding terms causes the sequence to converge to a specific number.
  • If the partial sums approach a particular value as 'n' goes to infinity, the series converges.
  • If not, the series diverges.
For example, in the telescoping series provided, \(S_n\) simplifies significantly due to term cancellation, leaving only a few terms鈥攎aking it easy to compute and thus find if the series converges.
Limit of a sequence
The limit of a sequence defines the value that the terms of a sequence "approach" as the sequence progresses indefinitely. For sequences derived from series, such as the sequence of partial sums, determining this limit helps in understanding the convergence behavior.In our telescoping series example, we consider the limit \(\lim_{n \rightarrow \infty} S_n\). Evaluating this limit will reveal the series' value:
  • The provided series transformed using partial fraction decomposition and telescoping simplifies to \(S_n = \frac{1}{7}\left(\frac{1}{5} - \frac{1}{2n+1}\right)\).
  • To find the limit, examine what happens as \(n\) becomes very large.
Here, the term \(\frac{1}{2n+1}\) becomes very small, approaching zero. Hence, the sequence of partial sums tends towards \(\frac{1}{35}\). That is why we conclude the series converges to \(\frac{1}{35}\), rather than diverging.
Partial fraction decomposition
Partial fraction decomposition is a technique used to break down rational functions into simpler fractions, making them easier to integrate or sum up in series. Often, it turns complex fractions into a series of simpler fractions, which can then be easily rearranged or cancelled out.In our telescoping series example, \(\frac{2}{(2k-1)(2k+1)}\) was expressed as:\[\frac{2}{(2k-1)(2k+1)} = \frac{A}{2k-1} + \frac{B}{2k+1}\]Here, the coefficients \(A\) and \(B\) were determined to be fraction values satisfying the original equation. Once these values were found:
  • The decomposition allowed us to find a telescoping series form, \(\frac{1}{7}\left(\frac{1}{2k-1} - \frac{1}{2k+1}\right)\), where terms neatly cancel out when summed.
  • This form simplifies solving the series and highlighting the meaning of each part, focusing on behavior as terms accumulate.
Thus, partial fraction decomposition acts as a crucial tool in simplifying and evaluating the given series.

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Most popular questions from this chapter

Evaluate the limit of the following sequences. $$a_{n}=\tan ^{-1}\left(\frac{10 n}{10 n+4}\right)$$

Consider the following situations that generate a sequence. a. Write out the first five terms of the sequence. b. Find an explicit formula for the terms of the sequence. c. Find a recurrence relation that generates the sequence. d. Using a calculator or a graphing utility, estimate the limit of the sequence or state that it does not exist. When a biologist begins a study, a colony of prairie dogs has a population of \(250 .\) Regular measurements reveal that each month the prairie dog population increases by \(3 \%\) Let \(p_{n}\) be the population (rounded to whole numbers) at the end of the \(n\) th month, where the initial population is \(p_{0}=250\).

Convergence parameter Find the values of the parameter \(p>0\) for which the following series converge. $$\sum_{k=2}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{p}}$$

Infinite products An infinite product \(P=a_{1} a_{2} a_{3} \ldots,\) which is denoted \(\prod_{k=1}^{\infty} a_{k}\) is the limit of the sequence of partial products \(\left\\{a_{1}, a_{1} a_{2}, a_{1} a_{2} a_{3}, \dots\right\\}\) a. Show that the infinite product converges (which means its sequence of partial products converges) provided the series \(\sum_{k=1}^{\infty} \ln a_{k}\) converges. b. Consider the infinite product $$P=\prod_{k=2}^{\infty}\left(1-\frac{1}{k^{2}}\right)=\frac{3}{4} \cdot \frac{8}{9} \cdot \frac{15}{16} \cdot \frac{24}{25} \cdots$$ Write out the first few terms of the sequence of partial products, $$P_{n}=\prod_{k=2}^{n}\left(1-\frac{1}{k^{2}}\right)$$ (for example, \(P_{2}=\frac{3}{4}, P_{3}=\frac{2}{3}\) ). Write out enough terms to determine the value of the product, which is \(\lim _{n \rightarrow \infty} P_{n}\). c. Use the results of parts (a) and (b) to evaluate the series $$\sum_{k=2}^{\infty} \ln \left(1-\frac{1}{k^{2}}\right)$$

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} e^{k}}{(k+1) !}$$
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