Question

Let \( a, b, c \in \mathbb{R} \). Which of the following values of \( a, b, c \) do NOT result in the convergence of the series
\[ \sum_{n=3}^{\infty} \frac{a^n}{n^b (\log n)^c} ? \]

• A. \( |a| < 1, b \in \mathbb{R}, c \in \mathbb{R} \)
• B. \( a = 1, b > 1, c \in \mathbb{R} \)
• C. \( a = 1, b \geq 0, c < 1 \)
• D. \( a = -1, b \geq 0, c > 0 \)

Hint:

For series with exponential terms, convergence depends on the growth rate of the denominator. A slower growth (such as when \( b \geq 0 \) and \( c < 1 \)) may prevent convergence.

Answer 1

The Correct Option is C

Step 1: Understanding the convergence criteria.
The series \( \sum_{n=3}^{\infty} \frac{a^n}{n^b (\log n)^c} \) will converge based on the behavior of \( a \), \( b \), and \( c \). Specifically, for convergence, the powers of \( n \) and \( \log n \) must grow sufficiently fast to counterbalance the terms \( a^n \).

Step 2: Analyzing the options.
(A) \( |a| < 1, b \in \mathbb{R}, c \in \mathbb{R} \): This condition ensures convergence, as the exponential term decays and \( n^b (\log n)^c \) grows sufficiently.
(B) \( a = 1, b > 1, c \in \mathbb{R} \): This also results in convergence, as the decay of \( n^b \) and \( (\log n)^c \) is fast enough.
(C) \( a = 1, b \geq 0, c < 1 \): This condition does not guarantee convergence, as the growth of \( n^b (\log n)^c \) is not fast enough to ensure convergence when \( b \geq 0 \) and \( c < 1 \).
(D) \( a = -1, b \geq 0, c > 0 \): This condition leads to convergence, as the series behaves similarly to option (B).

Step 3: Conclusion.
The correct answer is (C) \( a = 1, b \geq 0, c < 1 \).