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Estimate the value of the following convergent series with an absolute error less than \(10^{-3}\). $$\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{5}}$$

Short Answer

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#tag_title#Step 4: Finding the smallest integer value of n#tag_content#Taking the fifth root of both sides of the inequality, we get: $$(n+1) > (\sqrt[5]{10^{3}})$$ $$n+1 > 10$$ $$n > 9$$ Since we're looking for the smallest integer value of n, we can choose n = 10. Therefore, 10 terms are required for the absolute error to be less than \(10^{-3}\).

Step by step solution

01

Understanding the Alternating Series Estimation Theorem

The Alternating Series Estimation Theorem states that if a series is alternating and its terms are decreasing, then the error after n terms is less than or equal to the next term in the series. In this case, we have an alternating series because of the \((-1)^k\) term. The terms are decreasing since \(k^{5}\) increases while k increases.
02

Setting up the condition for convergence

According to the problem, we need to find the partial sum such that the absolute error is less than \(10^{-3}\). Using the Alternating Series Estimation Theorem, we have: $$|\frac{(-1)^{n+1}}{(n+1)^{5}}| < 10^{-3}$$ Our goal is to find the appropriate value of n that satisfies this condition.
03

Solving for n

To solve for n, we can ignore the negative sign and focus on the inequality: $$(n+1)^{5} > 10^{3}$$ Now, find the smallest integer value of n that satisfies this inequality:

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Key Concepts

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

Alternating Series Estimation Theorem
The Alternating Series Estimation Theorem is a valuable tool for determining the accuracy of an approximation of an alternating series. An alternating series is one where the terms alternate in sign. For a series \[S = a_1 - a_2 + a_3 - a_4 + ext{...}\]this theorem provides a neat way to estimate its sum by showing that if the absolute value of the terms decreases monotonically (getting smaller and smaller) and the limit of the terms approaches zero, the series converges.

The beauty of the Alternating Series Estimation Theorem lies in its ability to reveal the estimation error. If you approximate the sum of a finite number of terms, the error in your estimate is less than or equal to the absolute value of the first omitted term. This simplifies things significantly as it allows you to control and predict the precision of your approximation.
  • Key condition: Terms must be decreasing in absolute value.
  • Convergence guarantee: The limit of the terms must trend towards zero.
  • Error bound: The error is less than or equal to the first omitted term.
Absolute Error
In mathematics, when estimating values, it’s crucial to understand the concept of absolute error, especially in the context of series approximations. Absolute error provides a measure of how far off a calculated approximation is from the actual value you are trying to determine.

For instance, if you have a convergent series and you want the sum of just a few terms to approximate the entire series, absolute error gives a clear indicator of this difference. It is calculated as the absolute value of the difference between the estimated sum and the true sum. If the absolute error is small, it means your approximation is close to the true value.
  • Absolute error formula: \[|S - S_n|\]where \(S\) is the true sum and \(S_n\) is the estimated sum of the series.
  • In practical terms, we often try to keep the absolute error below a certain threshold to ensure the approximation is sufficiently accurate.
  • This measure is vital in ensuring the reliability and precision of numerical estimations.
Inequality
Inequality plays a crucial role in estimating series, particularly when using theorems like the Alternating Series Estimation Theorem. To confirm that an error in approximation remains within a desired bound, inequalities are employed to determine how many terms should be summed.

For example, if an inequality is set up as \[(n+1)^5 > 10^3\]in the context of an alternating series, solving this inequality provides the smallest integer, \(n\), needed to ensure the absolute error is smaller than a specified tolerance, here being \(10^{-3}\).

Inequalities allow us to determine solutions efficiently and find parameters that satisfy particular conditions in mathematical problems.
  • Key function: Dictates how many terms are necessary by setting bounds.
  • Essential in assuring precision by limiting the error margin.
  • Solving inequalities often includes rearranging terms and extracting roots or powers to find critical values, like in our example where solving led to identifying how many terms are necessary.

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Most popular questions from this chapter

In \(1978,\) in an effort to reduce population growth, China instituted a policy that allows only one child per family. One unintended consequence has been that, because of a cultural bias toward sons, China now has many more young boys than girls. To solve this problem, some people have suggested replacing the one-child policy with a one-son policy: A family may have children until a boy is born. Suppose that the one-son policy were implemented and that natural birth rates remained the same (half boys and half girls). Using geometric series, compare the total number of children under the two policies.

The Fibonacci sequence \(\\{1,1,2,3,5,8,13, \ldots\\}\) is generated by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1\). a. It can be shown that the sequence of ratios of successive terms of the sequence \(\left\\{\frac{f_{n+1}}{f_{n}}\right\\}\) has a limit \(\varphi .\) Divide both sides of the recurrence relation by \(f_{n},\) take the limit as \(n \rightarrow \infty,\) and show that \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}}=\frac{1+\sqrt{5}}{2} \approx 1.618\). b. Show that \(\lim _{n \rightarrow \infty} \frac{f_{n-1}}{f_{n+1}}=1-\frac{1}{\varphi} \approx 0.382\). c. Now consider the harmonic series and group terms as follows: $$\sum_{k=1}^{\infty} \frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\left(\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)$$ $$+\left(\frac{1}{9}+\cdots+\frac{1}{13}\right)+\cdots$$ With the Fibonacci sequence in mind, show that $$\sum_{k=1}^{\infty} \frac{1}{k} \geq 1+\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{3}{8}+\frac{5}{13}+\cdots=1+\sum_{k=1}^{\infty} \frac{f_{k-1}}{f_{k+1}}.$$ d. Use part (b) to conclude that the harmonic series diverges. (Source: The College Mathematics Journal, 43, May 2012)

$$\text {Evaluate each series or state that it diverges.}$$ $$\sum_{k=1}^{\infty}\left(\sin ^{-1}(1 / k)-\sin ^{-1}(1 /(k+1))\right)$$

The prime numbers are those positive integers that are divisible by only 1 and themselves (for example, 2,3,5,7, 11,13, \(\ldots\) ). A celebrated theorem states that the sequence of prime numbers \(\left\\{p_{k}\right\\}\) satisfies \(\lim _{k \rightarrow \infty} p_{k} /(k \ln k)=1 .\) Show that \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) diverges, which implies that the series \(\sum_{k=1}^{\infty} \frac{1}{p_{k}}\) diverges.

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{3 n^{2}}{4 n^{2}+1}=\frac{3}{4}$$

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