91Ó°ÊÓ

Let's start with the given series \(\sum_{n=1}^{\infty} \frac{\ln n}{n^{2}}\). This is an infinite series, which entails adding an infinite number of terms together. Indeed, it combines logarithmic functions and power functions.

For this series, a suitable test to determine whether it converges or diverges would be the Limit Comparison Test, as it effectively handles general comparisons between two series. The Limit Comparison Test says that if the limit as n goes to infinity of the terms in the series divided by the terms in a known convergent or divergent series exists and is not zero, then the original series has the same behavior (convergence or divergence) as the known series. We can compare the given series to the known series \( \sum_{n=1}^{\infty} \frac{1}{n^{2}} \), which is a convergent p-series with p = 2. Therefore, for the Limit Comparison Test to be applicable, we need to find the limit as n approaches infinity. Let's denote \[a_{n}=\frac{\ln n}{n^{2}} \] and \[b_{n}=\frac{1}{n^{2}}\]. Then, we need to calculate \(\lim_{n\to\infty} \frac{a_{n}}{b_{n}}\).

Calculate the limit \(\lim_{n\to\infty} \frac{\frac{\ln n}{n^{2}}}{\frac{1}{n^{2}}}\), which simplifies to \(\lim_{n\to\infty} \ln n\). As \(n\) goes to infinity, \(\ln n\) also goes to infinity. Because the Limit Comparison Test requires the limit to exist and not be zero, the test is inconclusive in this case.

If the Limit Comparison Test fails, we need to resort to a different test. A good choice in this case is the Integral Test, which allows us to compare the convergence or divergence of an infinite series to the convergence or divergence of a related improper integral. To use the Integral Test, we need to check that the function \(f(x) = \frac{\ln x}{x^{2}}\) (which corresponds to our original series where \(n\) has been replaced with the continuous variable \(x\)) is positive, continuous, and decreasing for \(x \geq 1\).

The improper integral corresponding to our series is \(\int_{1}^{\infty} \frac{\ln x}{x^{2}} dx\). Since the Integral Test states that if the integral of a function is convergent, then the series is also convergent, we need to try to evaluate this integral. By using the method of integration by parts with \(u=\ln x\) and \(dv=\frac{1}{x^{2}} dx\), we obtain that the integral is convergent. Therefore, we can conclude that the original series is also convergent.