Problem 53 - Essential Calculus Early Transcendentals

Question

If \(\Sigma a_{n}\) is convergent and \(\Sigma b_{n}\) is divergent, show that the series \(\Sigma\left(a_{n}+b_{n}\right)\) is divergent. [Hint: Argue by contradiction. \(]\)

Step-by-Step Solution

Answer

The series \(\Sigma \left(a_{n} + b_{n}\right)\) is divergent due to contradiction.

Step 1: Understand the Given Information

We are given the series \(\Sigma a_{n}\) which is convergent and the series \(\Sigma b_{n}\) which is divergent. We need to show that the series \(\Sigma \left(a_{n} + b_{n}\right)\) is divergent.

Step 2: Assume the Opposite (Contradiction)

For contradiction, let's assume that the series \(\Sigma \left(a_{n} + b_{n}\right)\) is convergent. That means the sum \(S = \Sigma \left(a_{n} + b_{n}\right)\) exists and is a finite value.

Step 3: Break Down \(\Sigma \left(a_{n} + b_{n}\right)\)

Since \(S = \Sigma \left(a_{n} + b_{n}\right)\) is assumed to be convergent, it implies \(S = \Sigma a_{n} + \Sigma b_{n}\). We know \(\Sigma a_{n}\) is convergent, so let it be \(S_a\). Hence, \(S = S_a + \Sigma b_{n}\).

Step 4: Analyze the Implication

Since \(S_a\) is a finite value and \(S\) is also finite (by our assumption), it follows that \(\Sigma b_{n} = S - S_a\) must also be finite. This consequence implies that \(\Sigma b_{n}\) is convergent.

Step 5: Identify the Contradiction

We assumed \(\Sigma \left(a_{n} + b_{n}\right)\) is convergent, leading to \(\Sigma b_{n}\) being convergent, which contradicts our given information that \(\Sigma b_{n}\) is divergent.

Step 6: Draw Conclusion

Since assuming \(\Sigma \left(a_{n} + b_{n}\right)\) convergent leads to a contradiction, it must be the case that \(\Sigma \left(a_{n} + b_{n}\right)\) is divergent.

Key Concepts

Convergent Series

When we talk about convergent series, we refer to a sequence of numbers, or terms, whose sum approaches a specific finite value as more terms are added. Key indicators include:

• The partial sums of the series approach a limit. Think of these partial sums as adding the terms one by one and observing where the total heads as we add more terms.
• A convergent series must satisfy the condition that the limit of the sequence of partial sums exists and has a finite value.

For instance, a simple example of a convergent series is the geometric series: \[ \Sigma \left( \frac{1}{2^n} \right) = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \] This series converges to 2. Understanding the concept of convergence is crucial because it helps us distinguish between series that settle to a value and those that do not.
A series that fails to converge is usually termed divergent. We'll explore more on divergences next.

Divergent Series

A divergent series is one where the sequence of partial sums does not approach a final value. In simpler terms, the sums continue to increase or decrease without settling to specific number. Some features include:

• The series may increase without bound, meaning it can grow indefinitely larger.
• The series may oscillate without approaching a limit, swinging between values.

An example of a divergent series is the simple harmonic series:\[ \Sigma \left( \frac{1}{n} \right) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots \]This series increases indefinitely, hence it diverges.
It's essential to identify when a series diverges, especially when examining combinations of series, like in our problem. Despite some terms being positive or negative, if overall the sum doesn't settle, it diverges.
Now, let’s see how proof by contradiction can elegantly show the divergence of combined series.

Proof by Contradiction

Proof by contradiction is a logical method where we assume the opposite of what we want to demonstrate, then show that this assumption leads to an impossible or false conclusion. This concept is wieldy and potent when you want to show impossibility. Here's the typical strategy:

• Begin by assuming that the opposite of the statement you want to prove is true.
• Use this assumption to logically arrive at a contradiction.
• The presence of a contradiction implies that your assumption must be wrong, confirming your original assertion.

In the context of the exercise, we assumed that if the series \( \Sigma \left(a_n + b_n\right) \) were convergent, it would imply \[ S = \Sigma a_n + \Sigma b_n \] was finite. However, since \( \Sigma b_n \) is already given as divergent, this situation creates a contradiction. Hence, the series \( \Sigma \left(a_n + b_n\right) \) cannot converge and is, therefore, divergent.
Utilizing proof by contradiction can help reinforce one's understanding as it allows exploring both the direct and inverse outcomes of mathematical scenarios.