Problem 34
Question
The partial sum indicated is used to estimate the sum of the series. Estimate the error. \(\sum_{k=1}^{\infty}(-1)^{k} \frac{1}{(10)^{k}} ; \quad s_{4}\).
Step-by-Step Solution
Verified Answer
The estimated error in the partial sum \( s_{4} \) is 0.00001.
1Step 1: Identify the Series
The series given is an alternating infinite series denoted by \(\sum_{k=1}^{\infty}(-1)^{k} \frac{1}{(10)^{k}}\). This will continue indefinitely, but since the sum \(\ s_{4} \) is mentioned, the calculation will focus on the first four terms of the series.
2Step 2: Calculate the Partial Sum
The partial sum be calculated by adding together the first four terms of the series, i.e., when \( k = 1 \), \( k = 2 \), \( k = 3 \), and \( k = 4 \). This gives the partial sum \(\ s_{4} = -\frac{1}{10} + \frac{1}{100} - \frac{1}{1000} + \frac{1}{10000} \). This simplifies to \(\ s_{4} = -0.0999 \).
3Step 3: Calculate the First Ignored Term
The error is estimated to be less than or equal to the absolute value of the first ignored term. This term is the fifth term in the series, which can be calculated by setting \( k = 5 \) in the formula for the series. This gives \( a_5 = (-1)^{5} \frac{1}{(10)^{5}} = -\frac{1}{100000} = -0.00001 \). The absolute value of this term is \( |a_5| = 0.00001 \).
4Step 4: Estimate the Error
As said before, the error in the estimate is less than or equal to the absolute value of \( a_5 \), which is \( 0.00001 \). Thus, the estimated error in the sum is \( 0.00001 \).
Key Concepts
Partial SumError EstimationInfinite Series
Partial Sum
A partial sum is a sum of a certain number of first terms of a series. In mathematical terms, it gives a way to approximate the sum of an infinite series by summing up a finite part of it. For example, in the series \( \sum_{k=1}^{\infty}(-1)^{k} \frac{1}{(10)^{k}} \), the partial sum \( s_4 \) is the sum of the first four terms:
- For \( k=1 \): \( -\frac{1}{10} = -0.1 \)
- For \( k=2 \): \( \frac{1}{100} = 0.01 \)
- For \( k=3 \): \( -\frac{1}{1000} = -0.001 \)
- For \( k=4 \): \( \frac{1}{10000} = 0.0001 \)
Error Estimation
When calculating the partial sum of an infinite series, many times we want to know how close our estimate is to the actual sum of the series. We do this through error estimation.
In an alternating series, error estimation is simplified by using the first term of the series that is not included in the partial sum.
For the given series, the error is estimated using the fifth term, which hasn't been added to the partial sum \( s_4 \).
In an alternating series, error estimation is simplified by using the first term of the series that is not included in the partial sum.
For the given series, the error is estimated using the fifth term, which hasn't been added to the partial sum \( s_4 \).
- Calculate the first ignored term (\( a_5 \)): \( (-1)^{5} \frac{1}{(10)^{5}} = -0.00001 \)
- Take the absolute value: \( |a_5| = 0.00001 \)
Infinite Series
An infinite series is a sequence of numbers that continues indefinitely. It is expressed as the sum of all its terms. The series \( \sum_{k=1}^{\infty}(-1)^{k} \frac{1}{(10)^{k}} \) is an example of an infinite series which keeps going without ending.
Since it is alternating, each term in the series switches its sign. This means that the summation process involves both positive and negative contributions that can cancel each other out, creating a more balanced approach to summing infinitely many numbers.
Infinite series can converge or diverge:
Since it is alternating, each term in the series switches its sign. This means that the summation process involves both positive and negative contributions that can cancel each other out, creating a more balanced approach to summing infinitely many numbers.
Infinite series can converge or diverge:
- **Convergent Series:** These series approach a specific value as more terms are added.
- **Divergent Series:** These series do not settle into a specific sum, possibly growing indefinitely.
Other exercises in this chapter
Problem 34
Estimate to within 0.01 by using series. $$\int_{0}^{1} x^{4} e^{-x^{2}} d x$$
View solution Problem 34
Find the interval of convergence. $$\sum \frac{1}{(\ln k)^{k}}(x-1)^{k}$$
View solution Problem 35
(a) Assume that \(d_{k} \rightarrow 0\) and show that \(\sum_{k=1}^{x}\left(d_{k}-d_{k+1}\right)=e d_{1}\) (b) Sum the following series: (i) \(\sum_{i=1}^{\inft
View solution Problem 35
Use a CAS to determine the Taylor polynomial \(P_{6}\) in powers of \((x-1)\) for \(f(x)=\arctan x.\)
View solution