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Chapter 9: Problem 38

Prove that if \(a_{n} \geq 0, b_{n}>0, \lim _{n \rightarrow \infty} a_{n} / b_{n}=\infty,\) and \(\Sigma b_{n}\) diverges then \(\Sigma a_{n}\) diverges.

Short Answer

Expert verified
Given conditions imply \(\Sigma a_n\) diverges as it is larger than a divergent series.

Step by step solution

01

Understand the Given Conditions

We have a sequence \(a_n\) such that \(a_n \geq 0\) for all \(n\), and another sequence \(b_n\) where \(b_n > 0\). We are given that \(\lim_{n \to \infty} \frac{a_n}{b_n} = \infty\), which means that for very large \(n\), the sequence \(\frac{a_n}{b_n}\) grows indefinitely. Additionally, we know that the series \(\Sigma b_n\) diverges.
02

Use the Limit Condition

Since \(\lim_{n \to \infty} \frac{a_n}{b_n} = \infty\), for any large constant \(M > 0\), there exists a positive integer \(N\) such that for all \(n > N\), \(\frac{a_n}{b_n} > M\). This implies that \(a_n > M \cdot b_n\) for all \(n > N\).
03

Relate the Divergence of \(\Sigma b_n\) to \(\Sigma a_n\)

Since \(a_n > M \cdot b_n\) for all \(n > N\), the series \(\Sigma a_n\) must be larger than \(M \cdot \Sigma_{n=N+1}^{\infty} b_n\). Since we know \(\Sigma b_n\) diverges, \(\Sigma_{n=N+1}^{\infty} b_n\) diverges as well.
04

Conclusion

Since \(M > 0\) is a constant, multiplying a divergent series \(\Sigma_{n=N+1}^{\infty} b_n\) by \(M\) results in a series that also diverges. Therefore, the inequality \(\Sigma_{n=N+1}^{\infty} a_n > M \cdot \Sigma_{n=N+1}^{\infty} b_n\) guarantees that \(\Sigma a_n\) diverges. Consequently, we conclude that \(\Sigma a_n\) diverges.

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Key Concepts

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

Limit Comparison Test
The Limit Comparison Test is a handy tool for analyzing the convergence or divergence of series. It's particularly useful when you have two sequences, say \(a_n\) and \(b_n\), and you know some information about one of their series. The test compares the limit of their ratios:
  • If \(\lim_{n \to \infty} \frac{a_n}{b_n} = L\) and \(0 < L < \infty\), then \(\Sigma a_n\) and \(\Sigma b_n\) either both converge or both diverge.
  • If \(L = 0\) and \(\Sigma b_n\) converges, then \(\Sigma a_n\) converges.
  • If \(L = \infty\) and \(\Sigma b_n\) diverges, as in our original problem, \(\Sigma a_n\) must also diverge.
This test relies on the principle that if \(a_n\) behaves similarly to \(b_n\), their series should behave in related manners. Hence, in the given exercise, because the ratio of their terms \(\lim_{n \to \infty} \frac{a_n}{b_n} = \infty\) implies that \(a_n\) is much larger than \(b_n\) for large \(n\), we conclude \(\Sigma a_n\) must diverge if \(\Sigma b_n\) does.
Series Convergence
Series Convergence refers to the behavior of the sum of terms of a sequence. A series \(\Sigma a_n\) is said to converge if its sequence of partial sums approaches a finite limit as \(n\) tends to infinity. Mathematically,
  • For a series \(\Sigma a_n\), if \(\lim_{n \to \infty} S_n = S\), where \(S_n = \sum_{k=1}^{n} a_k\), the series converges to \(S\).
  • If no such finite limit exists, the series is said to diverge.
In our problem, we're given that \(\Sigma b_n\) diverges. This means that the sum \(\sum_{k=1}^{n} b_k\) grows without bound as \(n\) increases. Since the series \(\Sigma a_n\) is ultimately larger than a multiple of a series that diverges, it cannot converge to a finite sum and hence must also diverge.
Sequences and Series
Understanding sequences and series is crucial for mastering calculus and analysis. Here's a brief breakdown:
  • Sequence: A sequence is a list of numbers, \(a_1, a_2, a_3, \ldots\), that follows a specific rule or pattern. We often consider the behavior as the indices \(n\) go to infinity.
  • Series: When we add the terms of a sequence together, we get a series. For example, \(a_1 + a_2 + a_3 + \ldots = \Sigma a_n\).
In the context of the problem, each \(a_n\) in the sequence satisfies \(a_n > 0\) and \(b_n > 0\), which means we deal with non-negative sequences. The convergence or divergence of their series depends heavily on how these terms behave as \(n\) becomes very large.
In summary, sequences provide the building blocks of series, and understanding their properties helps in analyzing whether series converge or diverge.

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Most popular questions from this chapter

Find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. \(1+2 x+2^{2} x^{2}+2^{3} x^{3}+2^{4} x^{4}+\cdots\)

Find the terms through \(x^{5}\) in the Maclaurin series for \(f(x) .\) Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x)\). $$ f(x)=\left(1-x^{2}\right)^{2 / 3} $$

Find the third-order Maclaurin polynomial for \((1+x)^{3 / 2}\) and bound the error \(R_{3}(x)\) if \(-0.1 \leq x \leq 0\).

Plot on the same axes the given function along with the Maclaurin polynomials of orders \(1,2,3,\) and \(4 .\) $$ \sin (\ln (1+x)) $$

Find the Maclaurin polynomial of order \(n\) for \(f(x)=1 /(1-x) .\) Then use it with \(n=4\) to approximate each of the following. (a) \(f(0.1)\) (b) \(f(0.5)\) (c) \(f(0.9)\) (d) \(f(2)\)
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