Question

Consider the series \(\sum_{k=3}^{\infty} \frac{1}{k(\ln k)(\ln \ln k)^{p}},\) where \(p\) is a real number. a. For what values of \(p\) does this series converge? b. Which of the following series converges faster? Explain. $$\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{2}} \text { or } \sum_{k=3}^{\infty} \frac{1}{k(\ln k)(\ln \ln k)^{2}} ?$$

Short answer

Answer: The series converges for $p > 1$, and the second series, $\sum_{k=3}^{\infty} \frac{1}{k(\ln k)(\ln \ln k)^{2}}$, converges faster than the first series, $\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{2}}$.

Step-by-step solution

Understanding Integral Test

Integral test can be used if the function \(f(k) = \frac{1}{k(\ln k)(\ln \ln k)^{p}}\) is positive, continuous, and decreasing on the interval \([3, \infty)\). Since we are working with the natural logarithm, these conditions hold true for \(k\geq3\). Now, if the improper integral \(\int_3^{\infty} f(k) dk\) converges, the series also converges, and if the integral diverges, the series diverges.

Applying the Integral Test

We will evaluate the improper integral \(\int_3^{\infty} \frac{1}{k(\ln k)(\ln \ln k)^{p}} dk\). To solve this integral, apply substitution: let \(u = \ln k\), so \(du = \frac{1}{k} dk\). Now, we will integrate \(\int_3^{\infty} \frac{1}{k(\ln k)(\ln \ln k)^{p}} dk\) for the new variables: $$\int_{\ln 3}^{\infty} \frac{1}{u (\ln u)^{p}} du$$ Now, apply the substitution technique again: let \(w = \ln u\), so \(dw = \frac{1}{u} du\). Thus, the integral becomes: $$\int_{\ln(\ln 3)}^{\infty} \frac{1}{(\ln w)^{p}} dw$$ At this point, we have an improper integral of the form \(1/(\ln w)^p\) with respect to \(w\). The behavior of this integral will depend on the value of \(p\):

Convergence of the Integral

We will analyze the convergence of $$\int_{\ln(\ln 3)}^{\infty} \frac{1}{(\ln w)^{p}} dw.$$ Observe that, if \(p>1\), the integral converges because it is then a \(p\)-series and for \(p \leq 1\), the integral diverges. Therefore, the original series converges for \(p>1\) and diverges for \(p \leq 1\). For part a, the series converges for \(p > 1\).

Comparing the Convergence Rate of Two Series

We are asked to compare the series: $$\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{2}} \text { and } \sum_{k=3}^{\infty} \frac{1}{k(\ln k)(\ln \ln k)^{2}}$$ To determine which series converges faster, we can compare the denominators of the terms in the series. We can see that the term \((\ln \ln k)^{2}\) in the second series makes the denominator larger than the denominator of the first series. So, the term in the second series is smaller when compared with the corresponding term in the first series: $$\frac{1}{k(\ln k)(\ln \ln k)^{2}} < \frac{1}{k(\ln k)^{2}}$$ It means that the second series converges faster than the first series. So, the series $$\sum_{k=3}^{\infty} \frac{1}{k(\ln k)(\ln \ln k)^{2}}$$ converges faster. For part b, the second series, \(\sum_{k=3}^{\infty} \frac{1}{k(\ln k)(\ln \ln k)^{2}}\), converges faster than the first series, \(\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{2}}\).

Key concepts

Integral Test

The integral test is a powerful tool for determining the convergence of infinite series. Imagine you have a function, like our example with \(f(k) = \frac{1}{k(\ln k)(\ln \ln k)^{p}}\). The integral test states that if this function is positive, continuous, and decreasing for all \(k\geq 3\), and you evaluate the improper integral \(\int_3^{\infty} f(k) \, dk\), the convergence of the integral mirrors the convergence of the series.

• If the integral converges, so does the series.
• If the integral diverges, the series also diverges.

This is incredibly useful because integrals are often easier to evaluate or estimate than infinite sums.

Improper Integral

An improper integral is used here to check if our series converges. It involves integrating across an infinite interval or at a point where the function isn’t well-behaved. In our case, we examined the integral:\[\int_3^{\infty} \frac{1}{k(\ln k)(\ln \ln k)^{p}} \, dk\]Using substitution, we changed variables from \(k\) to another variable like \(u\) or \(w\), making it more manageable. Evaluating improper integrals often involves:

• Changing the variable to simplify the expression.
• Determining the behavior of the integral by checking convergence at infinity.

Here, it hinges on understanding how quickly the logarithmic terms grow as \(k\) becomes very large.

P-Series

A \(p\)-series is a series of the form \(\sum \frac{1}{n^{p}}\). The convergence of \(p\)-series depends entirely on the value of \(p\):

• If \(p > 1\), the series converges.
• If \(p \leq 1\), the series diverges.

In our problem, we ended up with an integral of the form \(\int \frac{1}{(\ln w)^p} \, dw\). Recognizing this as analogous to a \(p\)-series, we deduced that for \(p > 1\), the integral converges. Therefore, our series \(\sum_{k=3}^{\infty} \frac{1}{k(\ln k)(\ln \ln k)^{p}}\) converges when \(p > 1\). Understanding \(p\)-series can unlock the convergence secrets of many other complex series.

Convergence Rate Comparison

When comparing series, understanding which converges faster helps us see their growth behavior. For our problem, we compared:\[\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{2}} \quad \text{and} \quad \sum_{k=3}^{\infty}\frac{1}{k(\ln k)(\ln \ln k)^{2}}\]The key is to compare their terms. The second series has an extra factor \((\ln \ln k)^{2}\) in the denominator. This means its terms become smaller faster as \(k\) increases. Therefore, it converges more quickly than the first series.

• The larger the denominator of the terms in a series, the smaller the terms themselves.
• Smaller terms lead to faster convergence if they approach zero more rapidly.

Overall, this comparison helps determine which solution is more effective for practical applications where convergence speed is crucial.