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 \(\sum_{k=3}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{p}}\) converges when \(p>1\). The series \(\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{2}}\) converges faster than the series \(\sum_{k=3}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{2}}\).

Step-by-step solution

Identify the series and convergence test

We have the series \(\sum_{k=3}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{p}}\). To find the values of \(p\) for which this series converges, we will use the comparison test and integral test.

Check convergence for p=1 using integral test

Let's first consider the case when \(p=1\). The series becomes \(\sum_{k=3}^{\infty} \frac{1}{k \ln k \ln(\ln k)}\). To apply the integral test, consider the function \(f(x) = \frac{1}{x \ln x \ln(\ln x)}\). We check that \(f(x)\) is positive, continuous, and decreasing for \(x\geq 3\). Now we evaluate the integral \(\int_{3}^{\infty} \frac{1}{x \ln x \ln(\ln x)} dx\). To evaluate this integral, we can use substitution: let \(u=\ln x \Rightarrow du=\frac{1}{x}dx\). The integral becomes: $$ \int_{\ln(3)}^{\infty} \frac{1}{u \ln(u)} du $$ We apply another substitution: let \(v=\ln u \Rightarrow dv=\frac{1}{u}du\). The integral becomes: $$ \int_{\ln(\ln(3))}^{\infty} \frac{1}{v} dv $$ This integral evaluates to \(\ln v\) evaluated from \(\ln(\ln(3))\) to \(\infty\), which is infinite. Since the integral is infinite, the series diverges for \(p=1\).

Apply the comparison test for p ≠ 1

For \(p \neq 1\), we compare the series with \(\sum_{k=3}^{\infty} \frac{1}{k \ln k}\), which is a known divergent series. If \(p > 1\), then \((\ln \ln k)^{p} > (\ln \ln k)^1\). Hence, \(\frac{1}{k \ln k(\ln \ln k)^{p}} < \frac{1}{k \ln k}\) because the denominator of the first fraction is greater. Since the comparison series diverges, our series converges by the comparison test when \(p > 1\). If \(p<1\), then \((\ln \ln k)^{p} < (\ln \ln k)^1\). Hence, \(\frac{1}{k \ln k(\ln \ln k)^{p}} > \frac{1}{k \ln k}\). Since the comparison series diverges, our series also diverges when \(p<1\) by the comparison test.

Convergence condition for p

From the above analysis, the series converges only when \(p>1\).

Comparing the convergence of given series

We need to compare the convergence of \(\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{2}}\) and \(\sum_{k=3}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{2}}\). We already know from step 3 that both series converge, and their general terms are \(\frac{1}{k(\ln k)^2}\) and \(\frac{1}{k \ln k(\ln \ln k)^2}\). Between these two, since $$ \frac{1}{k \ln k(\ln \ln k)^{2}} > \frac{1}{k(\ln k)^{2}} $$ it follows that the series \(\sum_{k=3}^{\infty} \frac{1}{k \ln k (\ln \ln k)^{2}}\) converges slower than the series \(\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{2}}\) because the denominator has an additional term, causing the fraction to be larger. Hence, the series \(\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{2}}\) converges faster.

Key concepts

Comparison Test

The comparison test is a handy tool for determining the convergence of a series by comparing it with another series whose behavior is already known. The basic idea is to match your series with one that either converges or diverges.

• If the series you know converges and your series has terms that are less than or equal to the known series terms from some point onward, your series converges too.
• On the other hand, if the known series diverges and has terms greater than or equal to it from some point onward, your series will diverge as well.

In the exercise, we used the comparison test to analyze when the series develops convergence based on the parameter \( p \). By selecting a suitable comparison series like \( \sum_{k=3}^{\infty} \frac{1}{k \ln k} \), we successfully deduced how the original series behaves by examining whether it has larger or smaller terms than this divergent series, given different values of \( p \).

Integral Test

The integral test is another excellent method for determining the convergence of infinite series. This test is best suited for series whose terms resemble a function that is positive, continuous, and decreasing.
When using the integral test, the idea is to integrate the function from the lower limit of the series to infinity.

• If the integral of the function from this point converges, the series converges.
• If the integral diverges, so does the series.

In the exercise, we applied the integral test when examining the series with \( p = 1 \). By using appropriate substitutions, we simplified the integral to a form that can be evaluated. Unfortunately, the result was an infinite integral, indicating that the series diverges for \( p = 1 \). This insight helps us narrow down the values of \( p \) for which the series converges.

Logarithmic Functions

Logarithmic functions play a crucial role in analyzing series, especially when dealing with convergence. A logarithmic function, in its simplest form, appears as \( \ln(x) \), which specifies the power to which a number (usually \( e \)) must be raised to produce \( x \). In convergence tests, they often appear in the denominator of series terms, slowing their growth.
One important aspect is the logarithmic hierarchy in functions such as \( \ln(\ln(x)) \). Such nested logarithms appear frequently, like in the series from the exercise, making series terms diminish much more slowly. Consequently, recognizing and evaluating the impact of these logarithmic structures is key to understanding series properties, such as in our comparison and integral tests.

Divergent Series

A divergent series is one that does not have a finite sum. Identifying a series as diverging tells us it keeps adding up towards infinity without settling anywhere.
In the exercises, divergence surfaces as the crux of the problem for specific values of \( p \). By showing that certain integral evaluations lead to infinity, it proves divergence when \( p = 1 \). Furthermore, by using the comparison test, it was demonstrated that when \( p < 1 \), the original series behaves similarly to a known divergent series, indicating divergence.
Understanding these behaviors and applying suitable tests establishes that a series diverges, sparing unnecessary efforts to find accurate summation as they do not exist. It becomes a critical step in mastering infinite series and their convergence properties.