91Ó°ÊÓ

We want to show that the terms of the series decrease in magnitude, i.e., \(|\frac{\ln(k+1)}{(k+1)^2}| \le |\frac{\ln k}{k^2}|\). This can be done by calculating the derivative of the expression inside the absolute value and checking if it's negative for the given range. First, calculate the derivative. Let \(f(k) = \frac{\ln k}{k^2}\). Then, find the derivative of \(f(k)\) as follows: $$f'(k) = \frac{d}{dk} \frac{\ln k}{k^2}$$ Now, apply the quotient rule: $$f'(k) = \frac{(k^2)(\frac{1}{k}) - (\ln k)(2k)}{(k^2)^2} = \frac{k - 2k\ln k}{k^4}$$ Since \(k \ge 2\), we can observe that \(2k\ln k > k\). Therefore: $$f'(k) < 0$$ This implies that the terms of the series decrease in magnitude, satisfying the first condition of the Alternating Series Test.

We want to show that the limit of the series approaches zero, i.e., \(\lim_{k \to \infty} \frac{\ln k}{k^2} = 0\). To do this, we can use L'Hopital's Rule to evaluate the limit: $$\lim_{k \to \infty} \frac{\ln k}{k^2}$$ Apply L'Hopital's Rule, taking the derivative of the numerator and denominator: $$\lim_{k \to \infty} \frac{\frac{1}{k}}{2k} = \lim_{k \to \infty} \frac{1}{2k^2} = 0$$ Since the limit of the series approaches zero, the second condition of the Alternating Series Test is satisfied.

Since both conditions of the Alternating Series Test are satisfied, we can conclude that the given series converges: $$\sum_{k=2}^{\infty}(-1)^{k} \frac{\ln k}{k^{2}}$$ converges.