Question

Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \ln (n+1)}{n+1} $$

Short answer

The given series converges.

Step-by-step solution

Identify the series

The given series is an alternating series with general term \(\frac{(-1)^{n+1} \ln(n+1)}{n+1}\)

Apply the Alternating Series Test

The Alternating Series Test states that an alternating series converges if both of the following conditions hold: (1) the sequence of terms \(\abs{a_n} = \frac{\ln(n+1)}{n+1}\) decreases to zero as \(n\) goes to infinity, and (2) the sequence is eventually decreasing. We need to check these conditions for the series.

Verify that the terms go to zero

Taking the limit of the absolute value of the series terms as \(n\) goes to infinity: \(\lim_{n \to \infty} \abs{a_n} = \lim_{n \to \infty} \frac{\ln(n+1)}{n+1}\). This limit is zero because \(\ln(n+1)\) grows slower than linearly

Verify the sequence is decreasing

We can test whether the sequence is decreasing by taking the derivative of \(b_n = \frac{\ln(n+1)}{n+1}\) and checking its sign: \(b_n' = \frac{1 - \ln(n+1)}{(n+1)^2}\). The derivative is negative for \(n > e\), so the sequence is eventually decreasing.

Conclude about the convergence

Since both conditions of the Alternating Series Test are satisfied, we conclude that the series does converge.