Problem 30 - Thomas Calculus

Question

Determining Convergence or Divergence In Exercises \(17-44,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n^{2}}\right)^{n}$$

Step-by-Step Solution

Answer

The series converges.

Step 1: Analyze the series

We are given the series \(\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n^{2}}\right)^{n}\). Our task is to determine if this series converges or diverges.

Step 2: Examine the general term

The general term of the series is \(a_n = \left(\frac{1}{n}-\frac{1}{n^{2}}\right)^{n}\). For large \(n\), the term \(\frac{1}{n}\) dominates \(\frac{1}{n^2}\), so the general term resembles \(\left(\frac{1}{n}\right)^{n}\).

Step 3: Simplify the general term

Simplify the general term:\[\left(\frac{1}{n}-\frac{1}{n^2}\right)^{n} = \left(\frac{n-1}{n^2}\right)^{n} = \left(\frac{n-1}{n}\cdot\frac{1}{n}\right)^n = \left(1 - \frac{1}{n}\right)^n\cdot\left(\frac{1}{n}\right)^n.\]

Step 4: Find a limiting behavior of the general term

Notice that \(\left(1-\frac{1}{n}\right)^{n}\) converges to \(e^{-1}\) as \(n\to\infty\). Therefore, \(a_n = \left(1 - \frac{1}{n}\right)^n \left(\frac{1}{n}\right)^n \approx e^{-1}\left(\frac{1}{n}\right)^n\).

Step 5: Apply the Limit Comparison Test

Since \(\left(\frac{1}{n}\right)^n\) is similar to a geometric series with ratio \(\frac{1}{n}\) which is less than 1 for \(n \geq 2\), and since the limit of \(e^{-1}\left(\frac{1}{n}\right)^n\) as \(n \to \infty\) is 0, the series \(\sum_{n=1}^{\infty} a_n\) converges by the Limit Comparison Test with a convergent geometric series.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a powerful method used to determine the convergence of infinite series. It involves comparing a given series to a known benchmark series. If both series behave similarly, then they either both converge or both diverge.

To apply the test, you need two series: the one you're investigating, \(\sum a_n\), and a comparison series, \(\sum b_n\). Calculate the limit \(\lim_{{n \to \infty}} \frac{a_n}{b_n}\). If this limit exists and is a positive finite number, then both series have the same convergence behavior.

In our example, \(a_n = \left(1 - \frac{1}{n}\right)^n \left(\frac{1}{n}\right)^n\) is compared to the series \(b_n = \left(\frac{1}{n}\right)^n\). Through simplification, these terms show very similar behavior, allowing us to conclude convergence through the Limit Comparison Test with a known convergent series.

Geometric Series

Geometric series are a special type of series where each term is obtained by multiplying the previous one by a constant ratio \(r\). The general form of a geometric series is \(\sum_{n=0}^{\infty} ar^n\), where \(a\) is the first term.

Convergence of a geometric series is based on the value of \(r\):

• If \(|r| < 1\), the series converges.
• If \(|r| \geq 1\), the series diverges.

In the provided exercise, after simplification, \(a_n \) behaves similar to \(\left(\frac{1}{n}\right)^n\), mirroring the structure of a geometric series with an ever-decreasing ratio, indicating convergence.

General Term Simplification

Simplifying the general term in a series can reveal insights that are not immediately obvious. Finding a simpler form helps to identify the series' convergence behavior more easily.

In our case, the general term is given as \(a_n = \left(\frac{1}{n}-\frac{1}{n^2}\right)^{n}\). For large \(n\), simplifying this to \(\left(1 - \frac{1}{n}\right)^n \left(\frac{1}{n}\right)^n\) shows us a compound term whose \(n\)-th power results in two individual behaviors: \(\left(1 - \frac{1}{n}\right)^n\) converges to \(e^{-1}\), and \(\left(\frac{1}{n}\right)^n\) shrinks rapidly. These insights are crucial in establishing the grounds for using the Limit Comparison Test effectively.