Problem 50 - Essential Calculus Early Transcendentals

Question

Suppose that \(\sum_{n=1}^{\infty} a_{n}\left(a_{n} \neq 0\right)\) is known to be a convergent series. Prove that \(\sum_{n=1}^{\infty} 1 / a_{n}\) is a divergent series.

Step-by-Step Solution

Answer

The series \(\sum_{n=1}^{\infty} \frac{1}{a_n}\) diverges since \(\frac{1}{a_n}\) grows without bound.

Step 1: Understanding the Problem

We need to prove that if \(\sum_{n=1}^{\infty} a_n\) is a convergent series, then \(\sum_{n=1}^{\infty} \frac{1}{a_n}\) is divergent. The key condition is that each term \(a_n\) is non-zero.

Step 2: Properties of Convergent Series

If \(\sum_{n=1}^{\infty} a_n\) converges, it means the terms \(a_n\) must approach zero as \(n\) approaches infinity. However, \(a_n\) cannot be zero because individually each \(a_n eq 0\).

Step 3: Behavior of \(1/a_n\) as \(n\to\infty\)

If \(a_n \to 0\), then \(\frac{1}{a_n}\) will grow without bound, meaning that \(\frac{1}{a_n}\) diverges as \(n\) increases.

Step 4: Comparison Test for Divergence

We use the fact that if the terms \(\frac{1}{a_n}\) do not approach zero, then \(\sum_{n=1}^{\infty} \frac{1}{a_n}\) diverges by the Test for Divergence. Since \(a_n \to 0\), \(\frac{1}{a_n} \to \infty\), thus \(\frac{1}{a_n}\) does not approach zero.

Step 5: Conclusion

Thus, since \(\sum_{n=1}^{\infty} a_n\) converges and \(a_n \to 0\), it must be that \(\sum_{n=1}^{\infty} \frac{1}{a_n}\) diverges because \(\frac{1}{a_n}\) does not approach zero.

Key Concepts

Convergent Series

A convergent series is a special kind of infinite series. In a convergent series, as we keep adding up terms, the total sum gets closer and closer to a certain number, or limit. This means it "settles down" to a point and doesn't go off to infinity. For example, think of the series from the harmonic sequence:

• \[\sum_{n=1}^{\infty} \frac{1}{n^2}\] This series converges to \(\pi^2/6\), which is approximately 1.644.

For a series to be convergent, each term in the series should "shrink" or get smaller as you progress along the series. That means, if you were to take any term further along in the series, it should be smaller than the one before it until they get very close to zero. Without these conditions being satisfied, an infinite series would not sum to a finite number.

Divergent Series

While in convergent series the sum approaches a specific number, divergent series display different behavior. In these series, no matter how far you go, the total sum does not settle down to any limit. It might grow without bounds, or it may not settle to any particular value at all. This means the series does not have a finite sum.

• \[\sum_{n=1}^{\infty} \frac{1}{n}\] - This is a classic example of a divergent series, known as the harmonic series. It grows indefinitely as more terms are added.

When terms of a series do not get close to zero as you include more of them, the series tends to be divergent. If the sequence of terms does not have zero as a limit, the series cannot possibly settle down to a finite sum.

Comparison Test

The Comparison Test is a handy tool for determining if a series converges or diverges. It's a method of comparing a series of interest with another series that is already known to be convergent or divergent. Here's how it works in simple steps:

• Compare two series: \(\sum a_n\) and \(\sum b_n\).
• If \(0 \leq a_n \leq b_n\), and \(\sum b_n\) is a known convergent series, then \(\sum a_n\) also converges.
• On the flip side, if \(\sum b_n\) is known to be divergent and \(a_n \geq b_n\), then \(\sum a_n\) is divergent as well.

In the context of the discussed exercise, by applying the Comparison Test, once you observe that \(1/a_n\) compares unfavorably with divergent series, it helps conclude that \(\sum_{n=1}^{\infty} 1/a_n\) diverges. This is because, for all large \(n\), \(a_n\) approaches zero, and therefore \(1/a_n\) becomes increasingly large, meaning it doesn't move towards zero, reinforcing divergence.