87. Explain why or why not
Determine whether the following statements are true and give an explanation or counterexample.


d. If ∑ pᵏ diverges, then ∑ (p + 0.001)ᵏ diverges, for a fixed real number p.

Verified step by step guidance

1. Recall the behavior of geometric series: For a fixed real number \( r \), the series \( \sum r^k \) converges if and only if \( |r| < 1 \), and diverges otherwise.
2. Given that \( \sum p^k \) diverges, this means that \( |p| \geq 1 \). This is the key starting point for analyzing the series \( \sum (p + 0.001)^k \).
3. Consider the value of \( |p + 0.001| \). Since \( p \) is fixed, adding 0.001 shifts the base slightly. We need to determine if this new base still satisfies \( |p + 0.001| \geq 1 \) or not.
4. If \( |p + 0.001| \geq 1 \), then \( \sum (p + 0.001)^k \) also diverges by the geometric series test. However, if \( |p + 0.001| < 1 \), then \( \sum (p + 0.001)^k \) converges, providing a counterexample to the statement.
5. Therefore, to conclude whether the statement is true or false, analyze specific values of \( p \) where \( |p| = 1 \) and see how the small increment affects convergence or divergence.

Key Concepts

Geometric Series and Convergence Criteria

A geometric series ∑ r^k converges if and only if the absolute value of the common ratio |r| is less than 1. If |r| ≥ 1, the series diverges. This criterion is fundamental to analyzing series of the form ∑ p^k and ∑ (p + 0.001)^k.

Effect of Changing the Common Ratio on Series Convergence

Altering the common ratio by a small amount (e.g., adding 0.001) can change whether the series converges or diverges. Even a slight increase can push the ratio beyond the convergence boundary, affecting the series' behavior significantly.

Counterexamples in Series Convergence

To determine the truth of a statement about series convergence, constructing counterexamples is essential. For instance, if ∑ p^k diverges but ∑ (p + 0.001)^k converges, this disproves the statement. Counterexamples help clarify the limits of general claims.