Question

Find the values of the parameter \(p>0\) for which the following series converge. $$\sum_{k=2}^{\infty} \frac{1}{(\ln k)^{p}}$$

Short answer

Answer: The series converges for \(p > 1\).

Step-by-step solution

Determine the function f(x)

Begin by identifying the function \(f(x)\) that corresponds to the given series. In this case, we have: $$f(x) = \frac{1}{(\ln x)^{p}}$$

Check the conditions for the integral test

To apply the integral test, we need to ensure that \(f(x)\) is positive, continuous, and decreasing on the interval \([2, \infty)\). Since \(\ln x > 0\) for \(x > 1\), and p > 0, it follows that \(f(x)\) is positive on the interval \((1, \infty)\). As \(f(x)\) is a composite of the natural logarithm function and a power function, it is also continuous on the interval \((1, \infty)\). Finally, to confirm that \(f(x)\) is decreasing, we can take the derivative: $$f'(x) = -\frac{p}{x (\ln x)^{p + 1}}$$ Since \(x > 1\), \(p > 0\), and \(\ln x > 0\), we have \(f'(x) < 0\) on the interval \((1, \infty)\). This means that \(f(x)\) is decreasing on the interval \((1, \infty)\), and we can apply the integral test.

Calculate the integral

Now, we will compute the integral: $$\int_{2}^{\infty} \frac{1}{(\ln x)^{p}} dx$$ Make the substitution \(u = \ln x\) and \(du = \frac{1}{x} dx\), we have: $$\int_{\ln2}^{\infty} \frac{1}{u^{p}} du$$ Now we use the power rule for integration: $$\int u^{-p} du = \frac{u^{-p+1}}{-p+1} + C$$ Applying the limits, we have: $$\lim_{T\to\infty} \left[ \frac{u^{-p+1}}{-p+1} \right]_{\ln 2}^{T}$$ When \(p > 1\), the integral converges: $$\frac{1}{1-p}\left(\lim_{T\to\infty} {T^{-p+1}}-\left(\ln2\right)^{1-p}\right)=\frac{1}{1-p}\left(0-\left(\ln2\right)^{1-p}\right)$$ When \(p \le 1\), the integral diverges: $$\lim_{T\to\infty} \left[ \frac{u^{-p+1}}{-p+1} \right]_{\ln 2}^{T} = \infty$$

Determine the values of p for which the series converges

Since the integral test tells us that the series and the integral have the same convergence properties (either both converge or both diverge), we can use the results from Step 3 to determine the values of p for which the series converges. From our calculations, the series converges when \(p > 1\) and diverges otherwise. Therefore, the values of the parameter p for which the given series converges are \(p > 1\).