Question

A fast computer can sum one million terms per second of the divergent series \(\sum_{n=2}^{N} \frac{1}{n \ln n} .\) Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed 100 .

Short answer

It takes approximately \(e^{e^{99.6335}} / 10^6\) seconds to exceed a sum of 100.

Step-by-step solution

Recall the Integral Test Approximation

The Integral Test states that for a continuous, positive, decreasing function, the series \(\sum_{n=2}^{\infty} f(n)\) can be compared to the integral \(\int_{2}^{N} f(x)\,dx\). In this case, we use \(f(n) = \frac{1}{n \ln n}\). To approximate the sum \(\sum_{n=2}^{N} \frac{1}{n \ln n}\), we compute \(\int_{2}^{N} \frac{1}{x \ln x} \,dx\).

Evaluate the Integral

Set up the integral \(\int_{2}^{N} \frac{1}{x \ln x} \,dx\). Use the substitution method: let \(u = \ln x\), then \(du = \frac{1}{x} \,dx\). Therefore, the integral becomes \(\int \frac{1}{u} \,du\), which evaluates to \(\ln |u| + C\), or \(\ln |\ln x| + C\) after substituting back.

Apply Limits to the Integral

Apply the limits to the evaluated integral: \(\int_{2}^{N} \frac{1}{x \ln x} \,dx = [\ln \ln x]_{2}^{N} = \ln \ln N - \ln \ln 2\).

Set the Sum Greater than 100

We need \(\ln \ln N - \ln \ln 2 > 100\). Solve for \(N\): \(\ln \ln N > 100 + \ln \ln 2\). Calculate \(\ln \ln 2\) and approximate \(N\).

Solve for N

First, compute \(\ln \ln 2 \approx \ln(0.693) \approx -0.3665\). Then, \(\ln \ln N > 100 - 0.3665 = 99.6335\). Therefore, \(\ln N > e^{99.6335}\). Calculate \(N\) using exponentials: \(N = e^{e^{99.6335}}\).

Calculate Time

Given that the computer can sum one million terms per second, find how long it takes to compute \(N\) terms. The time \(T\) in seconds is \(T = \frac{N}{10^6}\).

Key concepts

Divergent Series

A divergent series is a sequence of numbers whose collective sum does not converge to a finite limit. Unlike a convergent series, where the sum approaches a specific value, a divergent series continuously grows without bound as more terms are added.
For example, the series given by \[ \sum_{n=2}^{N} \frac{1}{n \ln n} \\] illustrates divergence. Each term may appear to get smaller, yet the overall sum keeps increasing indefinitely, fundamentally because it lacks a finite bound.
In practical terms, understanding whether a series is divergent is crucial for various applications in mathematics and physics, where findings often depend on convergence or divergence characteristics. Being aware that such series do not settle at a fixed value helps us determine how they behave as they extend towards infinity.

Continuous Functions

Continuous functions play a central role in calculus and series evaluation, particularly when applying the integral test. A function \( f(x) \) is considered continuous if, for every point \( x \) in its domain, the value of \( f(x) \) doesn't have any sudden jumps or breaks.
In the scenario presented, we deal with the function \( f(x) = \frac{1}{x \ln x} \). This function is continuous over the interval starting from 2, which ensures that integration can effectively approximate the behavior of the series.
Crucially, to use the integral test, the function in question must also be positive and decreasing. This means the function consistently decreases in value as \( x \) increases from 2, allowing us to make accurate comparisons with the integral.

Substitution Method

The substitution method is a calculus technique used to simplify the integration process. It involves replacing complicated expressions with simpler terms to make the integral easier to evaluate.
In the step-by-step solution for our exercise, we used the substitution method to solve the integral:\[ \int_{2}^{N} \frac{1}{x \ln x} \,dx \]By setting \( u = \ln x \), we have \( du = \frac{1}{x} \,dx \). This change transforms the integral into \[ \int \frac{1}{u} \,du, \]which is straightforward to evaluate and results in \( \ln |u| + C \). Substituting \( u = \ln x \) back, it becomes \( \ln |\ln x| + C \).
This method effectively broke down a potentially complex integral into a manageable form, enabling deeper understanding and simpler computation.

Series Approximation

Series approximation is a powerful tool in calculus, used to estimate the sum of a series when it is infeasible to calculate every term individually. This is particularly important with infinite or very large series.
Through the integral test, we approximate the sum \( \sum_{n=2}^{N} \frac{1}{n \ln n} \) by evaluating the integral \( \int_{2}^{N} \frac{1}{x \ln x} \,dx \). By determining \[ \ln \ln N - \ln \ln 2 > 100 \]from the integral results, we can establish a minimum value of \( N \) that satisfies this inequality, leading to the estimation of terms needed for the desired sum.
This technique helps in predicting the sum's behavior without performing exhaustive calculations. It is an essential concept in calculus as it simplifies complex tasks, making it easier to gain insights from challenging mathematical problems.