Question

Test for convergence: $$\sum_{n=2}^{\infty} \frac{1}{n \ln \left(n^{3}\right)}$$

Short answer

The series diverges.

Step-by-step solution

Simplify the expression inside the series

The given series is \(\frac{1}{n \, \text{ln} (n^3)}\). We can simplify \(\text{ln} (n^3)\) using the logarithmic property \(\text{ln} (a^b) = b \, \text{ln} (a)\). Thus, \(\text{ln} (n^3) = 3 \, \text{ln} (n)\). The series becomes \(\frac{1}{n \, 3 \, \text{ln} (n)} = \frac{1}{3 n \, \text{ln} (n)}\).

Identify the convergence test to use

Since the series involves terms that are \(\frac{1}{n \, \text{ln} (n)}\), which resemble a variant of the Harmonic series, we can use the Integral Test. The Integral Test is applicable if \(f(n)\) is a continuous, positive, and decreasing function for \(n \geq 2\).

Apply the Integral Test

Consider the function \(f(x) = \frac{1}{3x \, \text{ln} (x)}\). To apply the Integral Test, we evaluate the improper integral: \(\text{\int}_{2}^{\infty} \frac{1}{3x \, \text{ln} (x)} \, dx\). This integral tests the behavior of the series. Factor out the constant: \(\frac{1}{3} \text{\int}_{2}^{\infty} \frac{1}{x \, \text{ln} (x)} \, dx\).

Evaluate the improper integral

To evaluate \(\text{\int}_{2}^{\infty} \frac{1}{x \, \text{ln} (x)} \, dx\), use the substitution \( u = \text{ln} (x)\), which implies \(\frac{du}{dx} = \frac{1}{x}\) or \(dx = x \, du\). When \(x \rightarrow \infty\), \(u \rightarrow \infty\); and when \(x = 2\), \(u = \text{ln} (2)\). The integral becomes \(\text{\int}_{\text{ln}(2)}^{\infty} \frac{1}{u} \, du\). This simplifies to \(\text{\int}_{\text{ln}(2)}^{\infty} \frac{1}{u} \, du \, = \, [\text{ln}|u|]_{\text{ln}(2)}^{\infty} = \lim_{b \rightarrow \infty} \text{ln}(b) - \text{ln}(\text{ln}(2))\), which diverges because \(\text{ln}(b) \rightarrow \infty\).

Conclusion based on the Integral Test

Since the integral \(\text{\int}_{2}^{\infty} \frac{1}{x \, \text{ln} (x)} \, dx\) diverges, by the Integral Test, the original series \(\frac{1}{3 n \, \text{ln} (n)}\) also diverges.

Key concepts

Integral Test

The Integral Test is a valuable tool in determining the convergence or divergence of an infinite series. To employ this test, we must ensure that the related function, denoted as \(f(x)\), meets certain criteria: it must be continuous, positive, and decreasing for all \(x\) values greater than some value \(N\).

Here's how it works: you convert the series into a corresponding improper integral and evaluate it over some interval from \(N\) to infinity. If this improper integral converges, then the series converges. Conversely, if the integral diverges, so does the series.

For example, given the series \(\sum_{n=2}^{\infty} \frac{1}{3n \text{ln}(n)}\), the related function \(f(x)\) is \(\frac{1}{3x \text{ln}(x)}\). One would then look at the integral \(\int_{2}^{\infty} \frac{1}{3x \text{ln}(x)} dx\) to determine the convergence of the original series.

Logarithmic Properties

Understanding logarithmic properties can simplify complex expressions in series convergence problems. One useful property is \(\text{ln}(a^b) = b \text{ln}(a)\). This helps in breaking down expressions involving logarithms.

For instance, in the series \(\frac{1}{n \text{ln}(n^3)}\), the logarithm can be simplified: \(\text{ln}(n^3)\) becomes \(3 \text{ln}(n)\). Hence, our term is simplified to \(\frac{1}{3n \text{ln}(n)}\). This makes it easier to identify and apply the appropriate convergence test.

Simplifying expressions using logarithmic properties can often transform a complex problem into a more straightforward one, making it easier to analyze and solve.

Improper Integral

An improper integral is an integral that has one or more infinite limits and/or has an integrand that approaches infinity at one or more points within the integration range. Evaluating these integrals requires special techniques, often involving limits.

Consider the improper integral \(\int_{2}^{\infty} \frac{1}{x \text{ln}(x)} dx\) used in the Integral Test example. The first step is to factor out constants if possible, here: \(\frac{1}{3} \int_{2}^{\infty} \frac{1}{x \text{ln}(x)} dx\). Next, employing a substitution, such as \(u = \text{ln}(x)\), allows us to transform the integral into a more manageable form.

With \(u = \text{ln}(x)\), the integral becomes \(\int_{\text{ln}(2)}^{\infty} \frac{1}{u} du\), simplifying to \([\text{ln}|u|]_{\text{ln}(2)}^{\infty}\). This evaluates to \(\text{ln}(b) - \text{ln}(\text{ln}(2))\), which diverges as \(b \to \infty\).

Such detailed stepwise handling of improper integrals is crucial for determining the behavior of corresponding series.