Question

The series $\sum_{n=2}^{\infty} a_n$, where the $n^{th}$ term is $a_n = \frac{1}{\log n}$ :

• A. is divergent
• B. is convergent to 1
• C. is convergent to log 2
• D. is convergent to 2

Hint:

Since $\log n$ grows extremely slowly, its reciprocal $\frac{1}{\log n}$ decreases very slowly, making the series diverge much faster than the standard harmonic series.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To determine the convergence or divergence of an infinite series, we can compare its terms with those of a known benchmark series using the Direct Comparison Test.

Step 2: Detailed Explanation:
Let us consider the terms of the given series for $n \ge 2$:
\[ a_n = \frac{1}{\log n} \]
We know from logarithmic properties that for all $n \ge 2$:
\[ \log n < n \]
Taking the reciprocal of both sides reverses the inequality:
\[ \frac{1}{\log n} > \frac{1}{n} \]
Let us compare our series with the classical harmonic series:
\[ \sum_{n=2}^{\infty} \frac{1}{n} \]
The harmonic series is a well-known divergent series ($p$-series with $p = 1$).
By the Direct Comparison Test, since $\frac{1}{\log n} > \frac{1}{n}$ for all $n \ge 2$ and the smaller series $\sum \frac{1}{n}$ diverges, the larger series $\sum \frac{1}{\log n}$ must also diverge.
Therefore, the series is divergent.

Step 3: Final Answer
The correct option is (A).