Question

Let \( a_1 = b_1 = 0 \), and for each \( n \geq 2 \), let \( a_n \) and \( b_n \) be real numbers given by \[ a_n = \sum_{m=2}^{n} (-1)^{m} m (\log(m))^m \] \[ b_n = \sum_{m=2}^{n} \frac{1}{(\log(m))^m}. \] Then which one of the following is TRUE about the sequences \( \{a_n\} \) and \( \{b_n\} \)?

• A. Both \( \{a_n\} \) and \( \{b_n\} \) are divergent
• B. \( \{a_n\} \) is convergent and \( \{b_n\} \) is divergent
• C. \( \{a_n\} \) is divergent and \( \{b_n\} \) is convergent
• D. Both \( \{a_n\} \) and \( \{b_n\} \) are convergent

Hint:

In sequences with rapidly growing terms, divergence is often the result, especially when terms do not approach zero or the sums oscillate.

Answer 1

The Correct Option is D

Step 1: Analyzing the sequence \( \{a_n\} \).
The sequence \( a_n \) involves a sum of terms \( (-1)^m m (\log(m))^m \), where the magnitude of terms grows very quickly as \( m \) increases. The alternating sign and the large growth of \( m (\log(m))^m \) cause the sequence to oscillate and not settle to a limit. Hence, \( \{a_n\} \) diverges.

Step 2: Analyzing the sequence \( \{b_n\} \).
The sequence \( b_n \) involves the sum of terms \( \frac{1}{(\log(m))^m} \), where the denominator grows rapidly as \( m \) increases. Although each term becomes very small, the series does not tend to zero fast enough for the sequence to converge, meaning that \( \{b_n\} \) diverges.

Step 3: Conclusion.
Both \( \{a_n\} \) and \( \{b_n\} \) are divergent, so the correct answer is (A).