Chapter 10: Problem 6
(a) Suppose \(b_{n}+c_{n}
Short Answer
Expert verified
(a) Both \(\Sigma b_{n}\) and \(\Sigma c_{n}\) converge; (b) At least one of \(\Sigma b_{n}\) or \(\Sigma c_{n}\) diverges.
Step by step solution
01
Understand the Given Inequality for (a)
We are told that \(b_{n} + c_{n} < a_{n}\) and \(\Sigma a_{n}\) is a convergent series. This implies that the series formed by \(b_{n} + c_{n}\) terms is dominated by a convergent series.
02
Apply the Comparison Test for (a)
Since \(b_{n} + c_{n} < a_{n}\) and \(\Sigma a_{n}\) converges, by the comparison test, \(\Sigma(b_{n} + c_{n})\) also converges. Consequently, both \(\Sigma b_{n}\) and \(\Sigma c_{n}\), being non-negative series, also converge by the properties of series.
03
Understand the Given Inequality for (b)
For part (b), we are given \(a_{n} < b_{n} + c_{n}\) and \(\Sigma a_{n}\) diverges. This indicates that the partial sums of \(a_{n}\) grow without bound, and \(b_{n} + c_{n}\) must grow even faster.
04
Apply the Divergence Implication for (b)
Since \(a_{n} < b_{n} + c_{n}\) and the series \(\Sigma a_{n}\) diverges, by the comparison test, \(\Sigma(b_{n} + c_{n})\) must also diverge. Because \(b_{n}\) and \(c_{n}\) are positive, at least one of \(\Sigma b_{n}\) or \(\Sigma c_{n}\) must diverge as well.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Series Convergence
Series convergence is a fundamental concept in calculus and analysis concerning the behavior of infinite sums. When you have a sequence of terms, say \(a_n\), the series \(\Sigma a_n\) is essentially the sum of all these terms. For the series to converge, the sum must approach a finite limit as you keep adding terms indefinitely.
In mathematical terms, we say a series converges if its sequence of partial sums \(S_n = a_1 + a_2 + \ldots + a_n\) has a limit as \(n\) approaches infinity. If a series has no such limit, it diverges.
In mathematical terms, we say a series converges if its sequence of partial sums \(S_n = a_1 + a_2 + \ldots + a_n\) has a limit as \(n\) approaches infinity. If a series has no such limit, it diverges.
- Convergence implies stability—in essence, you can predict what happens when many elements of the series are combined.
- Checking convergence often involves using tests like the Comparison Test, Ratio Test, or Integral Test.
Comparison Test
The Comparison Test is a handy tool used to determine the convergence or divergence of infinite series. This test is based on comparing a given series with another one whose convergence properties are known.
To use the comparison test, you need two series: \(\Sigma a_n\) and \(\Sigma b_n\). You compare their terms, and if \(0 \leq a_n \leq b_n\) for all \(n\) from some point onwards, you can make certain inferences:
To use the comparison test, you need two series: \(\Sigma a_n\) and \(\Sigma b_n\). You compare their terms, and if \(0 \leq a_n \leq b_n\) for all \(n\) from some point onwards, you can make certain inferences:
- If \(\Sigma b_n\) converges, then \(\Sigma a_n\) also converges.
- If \(\Sigma a_n\) diverges, then \(\Sigma b_n\) also diverges.
Divergence of Series
The divergence of a series means that the series does not converge to any finite limit. This happens when the sequence of partial sums goes to infinity or does not settle at a stable value. Understanding divergence requires a solid grasp of infinite series and an ability to recognize when partial sums show no signs of reaching a finite value.
For example, a simple divergent series is the harmonic series \(\Sigma \frac{1}{n}\), where the sum just grows larger without bound as more terms are added.
For example, a simple divergent series is the harmonic series \(\Sigma \frac{1}{n}\), where the sum just grows larger without bound as more terms are added.
- Divergence implies unpredictability; unlike convergent series, divergent ones don't lead to a nice, finite answer.
- Determining divergence often uses tests similar to convergence tests, like the Comparison Test, where diverging series can set a benchmark.