Chapter 11: Problem 82
Suppose that \(\Sigma_{n-1}^{\infty} a_{n}\left(a_{n} \neq 0\right)\) is known to be a convergent series. Prove that \(\Sigma_{n-1}^{\infty} 1 / a_{n}\) is a divergent series.
Short Answer
Expert verified
\(\sum_{n=1}^{\infty} 1/a_n\) is divergent because \(1/a_n\) does not approach zero.
Step by step solution
01
Understanding Series Convergence
Given that the series \(\sum_{n=1}^{\infty} a_n\) is convergent, this implies that the terms \(a_n\) approach 0 as \(n\) approaches infinity. This is a necessary condition for the convergence of any series. Specifically, \(\lim_{n \to \infty} a_n = 0\).
02
Revisiting the Theory of Convergence and Divergence
Recall that if a series \(\sum_{n=1}^{\infty} b_n\) is convergent, the limit \(\lim_{n \to \infty} b_n = 0\) must also hold. However, the converse is not true; just because \(b_n\) approaches 0 does not mean that \(\sum_{n=1}^{\infty} b_n\) converges. For the series \(\sum_{n=1}^{\infty} 1/a_n\) to be diverging, we will utilize this concept.
03
Exploring Behavior of \(\sum_{n=1}^{\infty} 1/a_n\)
Since \(\lim_{n \to \infty} a_n = 0\), the terms \(1/a_n\) do not approach zero. In fact, as \(a_n\) gets smaller, \(1/a_n\) becomes larger. This indicates that the terms \(1/a_n\) do not meet the necessary condition for the convergence of a series.
04
Applying the Divergence Test
The divergence test states that if \(\lim_{n \to \infty} \frac{1}{a_n} eq 0\), then the series \(\sum_{n=1}^{\infty}\frac{1}{a_n}\) is divergent. Since we know \(\lim_{n \to \infty} a_n = 0\) and make \(\frac{1}{a_n}\) larger without bound, it is confirmed that \(\lim_{n \to \infty} \frac{1}{a_n} = \infty\), thus \(\sum_{n=1}^{\infty} \frac{1}{a_n}\) must be divergent.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Divergent Series
In mathematics, a divergent series is one whose terms do not approach a definite limit as you add more of them. Think of a series as a never-ending sum. In a divergent series, the sum either shoots up to infinity, plunges down to negative infinity, or oscillates indefinitely without settling at a particular value.
A classic example of a divergent series is the harmonic series: \[\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\]As you add more terms, the sum keeps growing, ultimately heading towards infinity. This infinite growth is what makes this series divergent.
A classic example of a divergent series is the harmonic series: \[\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\]As you add more terms, the sum keeps growing, ultimately heading towards infinity. This infinite growth is what makes this series divergent.
- In contrast to convergent series, a divergent series fails to meet the criteria for convergence.
- Divergence means adding more terms doesn't stabilize the sum to a fixed number.
Limit of a Sequence
The limit of a sequence refers to the number a sequence converges to as its terms progress towards infinity. A sequence \(a_n\) approaches a limit \(L\) if the terms get arbitrarily close to \(L\) as \(n\) becomes very large.
This can be mathematically represented as:\[\lim_{n \to \infty} a_n = L\]It's a fundamental concept used to evaluate series convergence, as it enables understanding how sequences behave at their extremes.
This can be mathematically represented as:\[\lim_{n \to \infty} a_n = L\]It's a fundamental concept used to evaluate series convergence, as it enables understanding how sequences behave at their extremes.
- A sequence with a well-defined limit is classified as convergent.
- If it doesn't approach any specific value, it's called divergent.
Necessary Condition for Convergence
Before declaring a series as convergent, it must satisfy specific preconditions, one being the necessary condition for convergence. This rule states that for a series \(\sum_{n=1}^{\infty} a_n\) to converge, its terms \(a_n\) must approach zero:
\[\lim_{n \to \infty} a_n = 0\]This is a crucial first check when examining the potential convergence of a series.
\[\lim_{n \to \infty} a_n = 0\]This is a crucial first check when examining the potential convergence of a series.
- This condition arises because if \(a_n\) doesn't approach zero, adding infinitely many non-zero terms can only lead to divergence.
- Despite being necessary, this condition alone isn't sufficient; other characteristics must be considered to confirm convergence.
Divergence Test
The divergence test is a basic yet powerful tool to establish whether a series is divergent. It's a quick way to identify series that are not convergent. The divergence test states:
If \(\lim_{n \to \infty} a_n eq 0\), then the series \(\sum_{n=1}^{\infty} a_n\) is divergent.
If \(\lim_{n \to \infty} a_n eq 0\), then the series \(\sum_{n=1}^{\infty} a_n\) is divergent.
- This test helps rule out series as convergent if their terms don't decline to zero.
- However, this test's reverse isn't true; terms approaching zero don't guarantee convergence.