Problem 45 - Essential Calculus

Question

Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=2}^{\infty} \frac{\ln n}{n+1} $$

Step-by-Step Solution

Answer

By using the Direct Comparison Test and comparing the given series to the harmonic series, we can conclude that the given series \( \sum_{n=2}^{\infty} \frac{\ln n}{n+1} \) is divergent.

Step 1: Understand the Direct Comparison Test

The Direct Comparison Test is a method to determine if an infinite series converges or diverges. The idea is that if series a_n is known to converge or diverge, then a comparison series b_n can be found such that \(0 \leq a_n \leq b_n\) or \(0 \leq b_n \leq a_n \) for all n. If a_n converges and \(a_n \leq b_n\) for all n, then b_n also converges. If a_n diverges and \(b_n \leq a_n\) for all n, then b_n also diverges.

Step 2: Establish a Comparison Series

The harmonic series \( \sum_{n=2}^{\infty} \frac {1}{n} \) is a known divergent series which seems like it could be used for comparison with the original series \( \sum_{n=2}^{\infty} \frac{\ln n}{n+1} \). Let's evaluate and compare the terms of these series for large n.

Step 3: Comparing terms of the series

Consider the inequality \(0 \leq \frac {1}{n} \leq \frac {\ln n}{n+1}\) for \(n \geq 2\). We can see that both inequalities are true: the term on the left is always positive and as n goes to infinity, the fractions are decreasing. Using the Direct Comparison Test, since the harmonic series \( \frac {1}{n} \) is diverging, the original series \( \frac {\ln n}{n+1} \) will also diverge because its terms are greater than the terms of a known diverging series.