Problem 29
Question
For the series given, determine how large \(n\) must be so that using the nth partial sum to approximate the series gives an error of no more than 0.0002. $$ \sum_{k=1}^{\infty} \frac{1}{1+k^{2}} $$
Step-by-Step Solution
Verified Answer
\( n = 5000 \)
1Step 1: Understanding the Series
The series under consideration is \( \sum_{k=1}^{\infty} \frac{1}{1+k^{2}} \). We need to find \( n \) such that the error in the \( n \)-th partial sum is no greater than 0.0002. This means the remainder (or tail) of the series, \( R_n \), which is the sum of terms from \( n+1 \) onward, must be less than or equal to 0.0002.
2Step 2: Estimating the Tail of the Series
For large \( k \), \( \frac{1}{1+k^2} \) is approximately similar to \( \frac{1}{k^2} \), which suggests using the comparison test with the series \( \sum_{k=1}^{\infty} \frac{1}{k^{2}} \), a convergent p-series with \( p=2 \). The remainder \( R_n = \sum_{k=n+1}^{\infty} \frac{1}{1+k^{2}} \) should therefore be compared with \( \frac{1}{n} \), which is a known result for the error bound of the similar \( \frac{1}{k^2} \).
3Step 3: Applying the Error Bound
We need \( R_n \leq 0.0002 \). To use the error bound \( R_n \approx \frac{1}{n} \), we solve the inequality: \( \frac{1}{n} \leq 0.0002 \). This simplifies to \( n \geq \frac{1}{0.0002} = 5000 \). Hence, \( n \) should be at least 5000.
Key Concepts
Error BoundConvergent SeriesComparison Testp-series
Error Bound
In the context of infinite series, the error bound is a crucial concept. It helps us determine how accurately a partial sum estimates the entire series. When approximating a series using its partial sums, the error bound gives an upper limit on the error that might occur.
Consider an infinite series \( \sum_{k=1}^{\infty} a_k \). If we stop at the \(n\)-th term, the partial sum is \( S_n = \sum_{k=1}^{n} a_k \). The error, represented by \( R_n \), is the sum of all remaining terms to infinity from \( n+1 \) onwards:
This means our partial sum approximates the entire sum to within a fraction of that small number.
For the series \( \sum_{k=1}^{\infty} \frac{1}{1+k^{2}} \), the task is to ensure \( R_n \leq 0.0002 \). Therefore, knowing how to calculate or estimate \( R_n \) is vital. Often, it involves comparing \( R_n \) to simpler convergent sums or known inequalities.
Consider an infinite series \( \sum_{k=1}^{\infty} a_k \). If we stop at the \(n\)-th term, the partial sum is \( S_n = \sum_{k=1}^{n} a_k \). The error, represented by \( R_n \), is the sum of all remaining terms to infinity from \( n+1 \) onwards:
- \( R_n = \sum_{k=n+1}^{\infty} a_k \)
This means our partial sum approximates the entire sum to within a fraction of that small number.
For the series \( \sum_{k=1}^{\infty} \frac{1}{1+k^{2}} \), the task is to ensure \( R_n \leq 0.0002 \). Therefore, knowing how to calculate or estimate \( R_n \) is vital. Often, it involves comparing \( R_n \) to simpler convergent sums or known inequalities.
Convergent Series
When discussing series like \( \sum_{k=1}^{\infty} \frac{1}{1+k^{2}} \), one important question is whether the series converges. A series is said to converge if its sequence of partial sums approaches a definite number as more terms are added.
For a series to be effectively used or analyzed, convergence is key.
An example of a convergent series is the \( p \)-series given by \( \sum_{k=1}^{\infty} \frac{1}{k^p} \), which converges for \( p > 1 \).
The series \( \sum_{k=1}^{\infty} \frac{1}{1+k^{2}} \) is similar to a \( p \)-series with \( p = 2 \), thus guaranteeing its convergence. This ensures that the errors in approximating such a series using partial sums can be made arbitrarily small.
For a series to be effectively used or analyzed, convergence is key.
- A series \( \sum_{k=1}^{\infty} a_k \) converges if, as \( n \to \infty \), the partial sum \( S_n \) approaches a finite limit \( L \).
An example of a convergent series is the \( p \)-series given by \( \sum_{k=1}^{\infty} \frac{1}{k^p} \), which converges for \( p > 1 \).
The series \( \sum_{k=1}^{\infty} \frac{1}{1+k^{2}} \) is similar to a \( p \)-series with \( p = 2 \), thus guaranteeing its convergence. This ensures that the errors in approximating such a series using partial sums can be made arbitrarily small.
Comparison Test
The comparison test is a practical method for determining the convergence of series by comparing them to known benchmarks.
If we suspect a series is convergent, we can find another series with a similar structure that we already know converges or diverges.
For our series \( \sum_{k=1}^{\infty} \frac{1}{1+k^{2}} \), the comparison can be made to the \( p \)-series \( \sum_{k=1}^{\infty} \frac{1}{k^{2}} \), which is known to converge.
If we suspect a series is convergent, we can find another series with a similar structure that we already know converges or diverges.
For our series \( \sum_{k=1}^{\infty} \frac{1}{1+k^{2}} \), the comparison can be made to the \( p \)-series \( \sum_{k=1}^{\infty} \frac{1}{k^{2}} \), which is known to converge.
- To use the comparison test, partially sum the series being analyzed and the comparison series.
- Compare the terms: if \( a_k \leq b_k \) and \( \sum b_k \) converges, then \( \sum a_k \) also converges.
p-series
A \( p \)-series is a fundamental type of mathematical series, given by \( \sum_{k=1}^{\infty} \frac{1}{k^p} \), where \( p \) is a positive constant. The convergence of a \( p \)-series is well understood and provides a general tool for analyzing other series.
For the series \( \sum_{k=1}^{\infty} \frac{1}{1+k^{2}} \), we observe similarities to the \( \sum_{k=1}^{\infty} \frac{1}{k^2} \), a convergent \( p \)-series with \( p = 2 \).
Using these known characteristics, we can strategically apply convergence tests and error estimates derived from \( p \)-series to ensure partial sums accurately approximate infinite series.
This approach showcases the importance of recognizing and leveraging classic mathematical constructs like \( p \)-series in complex series analysis.
- If \( p > 1 \), the \( p \)-series converges; if \( p \leq 1 \), the series diverges.
For the series \( \sum_{k=1}^{\infty} \frac{1}{1+k^{2}} \), we observe similarities to the \( \sum_{k=1}^{\infty} \frac{1}{k^2} \), a convergent \( p \)-series with \( p = 2 \).
Using these known characteristics, we can strategically apply convergence tests and error estimates derived from \( p \)-series to ensure partial sums accurately approximate infinite series.
This approach showcases the importance of recognizing and leveraging classic mathematical constructs like \( p \)-series in complex series analysis.
Other exercises in this chapter
Problem 29
Determine convergence or divergence for each of the series. Indicate the test you use. \(\sum_{n=1}^{\infty} \frac{4+\cos n}{n^{3}}\)
View solution Problem 29
Classify each series as absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty} \frac{(-3)^{n+1}}{n^{2}} $$
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Find an explicit formula a \(a_{n}=\) ____ for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \right
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Calculate, accurate to five decimal places, $$ \int_{0}^{0.5} \sin \sqrt{x} d x $$
View solution