Problem 85 - Calculus: Early Transcendentals

Question

If \( \sum a_n \) is convergent and \( \sum b_n \) is divergent, show that the series \( \sum \left( a_n + b_n \right) \) is divergent. [Hint: Argue by contradiction.]

Step-by-Step Solution

Answer

The series \( \sum (a_n + b_n) \) is divergent.

Step 1: Understand the Given Information

We are given two series, \( \sum a_n \) which is convergent, and \( \sum b_n \) which is divergent. We need to analyze the convergence or divergence of the series \( \sum (a_n + b_n) \).

Step 2: Recall the Definitions

Recall that a series \( \sum x_n \) is convergent if the sequence of partial sums \( S_n = x_1 + x_2 + ... + x_n \) approaches a finite limit as \( n \to \infty \). A series is divergent if its sequence of partial sums does not approach a finite limit.

Step 3: Assume for Contradiction

Assume, for the sake of contradiction, that \( \sum (a_n + b_n) \) is convergent. This means the sequence of partial sums \( T_n = (a_1 + b_1) + (a_2 + b_2) + ... + (a_n + b_n) \) approaches a finite limit as \( n \to \infty \).

Step 4: Write the Partial Sum Equation

The partial sum of the series \( \sum (a_n + b_n) \) is \( T_n = S_n + U_n \), where \( S_n = a_1 + a_2 + ... + a_n \) (partial sum of \( \sum a_n \)) and \( U_n = b_1 + b_2 + ... + b_n \) (partial sum of \( \sum b_n \)).

Step 5: Analyze the Implication

Since we assume \( T_n \) converges and since \( S_n \) converges (by hypothesis), \( U_n \) must also converge for \( T_n \) to converge. However, this contradicts the fact that \( \sum b_n \) is divergent (\( U_n \) must not converge).

Step 6: Conclude the Proof

The contradiction shows that our assumption that \( \sum (a_n + b_n) \) is convergent is false. Therefore, \( \sum (a_n + b_n) \) must be divergent.

Key Concepts

Understanding Series Convergence

In mathematics, the concept of series convergence is essential. The term refers to whether the series have a definitive or determinate sum when they continue indefinitely. Imagine adding an infinite series of numbers together; if these numbers add up to a finite number, then the series is said to "converge". For instance, a series \( \sum a_n \) is convergent if its partial sums \( S_n = a_1 + a_2 + \ldots + a_n \) approach a finite limit as \( n \to \infty \).
When you hear about series that converge, visualize a scenario where the total of the numbers eventually stalls and finds a sweet spot, no longer increasing significantly as more terms are added.
But not all series converge. Divergent series behave differently, where partial sums keep growing without ever settling down to a single, definitively finite value.

Exploring Partial Sums

The term "partial sums" is crucial when dealing with series. A partial sum is simply the sum of the first few terms in a series. Imagine a series like this: \( a_1 + a_2 + a_3 + \ldots \)
If you just add the first two terms \( a_1 \) and \( a_2 \), you've found the first partial sum \( S_2 \). Partial sums reflect how a series accumulates value as you progress through its list of terms.

• For a convergent series, the sequence of partial sums will always narrow down to a specific finite number.
• In contrast, for a divergent series, the partial sums will fail to settle at a firm number, shooting off towards infinity or oscillating endlessly.

Recognizing the behavior of partial sums helps in understanding the overall nature of the series, whether it converges or diverges. Without partial sums, it would be challenging to comprehend the infinite nature of series accurately.

The Contradiction Method

Using the contradiction method in mathematical proofs can be particularly powerful. It involves assuming the opposite of what you want to prove and then finding a logical fallacy or contradictory statement within that assumption.
In the case of series \( \sum(a_n + b_n) \), suppose it were assumed to be convergent. Assuming convergence means looking at the partial sums \( T_n = S_n + U_n \) (where \( S_n \) and \( U_n \) are the partial sums of \( \sum a_n \) and \( \sum b_n \), respectively).

• If \( T_n \) converges, then both \( S_n \) and \( U_n \) must generally converge.
• However, since \( \sum b_n \) is given as divergent, \( U_n \) should not converge.

This perceived convergence of \( T_n \) alongside the forced convergence of \( U_n \) contradicts the initial truth of \( \sum b_n \) diverging.
This contradiction tells us that our assumption is incorrect; the series \( \sum (a_n + b_n) \) must indeed be divergent. The contradiction method provides a path to clarity, stripping away false assumptions by showing where they fall apart logically.