91Ó°ÊÓ

We have the function \(f(x) = \frac{1}{(\ln x)^{p}}\). It's given that \(p > 0\), and thus \(f(x)\) is continuous, positive, and decreasing for \(x \geq 2\). We can now apply the Integral Test.

According to the Integral Test, the series converges if and only if the improper integral converges: $$\int_{2}^{\infty} \frac{1}{(\ln x)^{p}}\, dx$$

We will evaluate the improper integral using the substitution \(u = \ln x\), so \(du = \frac{1}{x} dx\). Then, the limits of integration change as well: for \(x=2\), we have \(u=\ln2\), and as \(x \to \infty\), \(u \to \infty\). Now we have: $$\int_{\ln 2}^{\infty} \frac{1}{u^{p}}(\frac{1}{x}) du = \int_{\ln 2}^{\infty} \frac{1}{u^{p}}\, du$$ The integral converges when \(p > 1\). We can justify this as follows: $$\int_{\ln 2}^{\infty} \frac{1}{u^{p}}\, du =\lim_{t\to \infty} \int_{\ln 2}^{t} \frac{1}{u^{p}}\, du$$ When \(p > 1\), the antiderivative of \(\frac{1}{u^{p}}\) is \(\frac{-1}{(p-1)u^{p-1}}\), so we have: $$\lim_{t\to \infty}\left(\frac{-1}{(p-1)(\ln 2)^{p-1}} - \frac{-1}{(p-1)t^{p-1}}\right) = \frac{1}{(p-1)(\ln 2)^{p-1}}$$ Now we can conclude that the series converges for \(p > 1\). Also, we know that for \(p\leq 1\), the integral will not converge based on the properties of functions of this type.

The series \(\sum_{k=2}^{\infty} \frac{1}{(\ln k)^{p}}\) converges if and only if \(p > 1\).