Problem 26 - Thomas Calculus

Question

In Exercises \(17-46,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$ \sum_{n=1}^{\infty}\left(1-\frac{1}{3 n}\right)^{n} $$

Step-by-Step Solution

Answer

The series diverges because its terms do not approach zero.

Step 1: Identify the Series Type

The given series is \( \sum_{n=1}^{\infty}\left(1-\frac{1}{3 n}\right)^{n} \). This form resembles a sequence with a complex exponential character, which suggests the use of limits for determining convergence.

Step 2: Evaluate the argument using exponential form

Recognize that \(\left(1-\frac{1}{3n}\right)^n \) resembles the form of the exponential function. Re-writing it using the natural exponential function gives: \((1-x)^n \approx e^{-nx}\), where \(x = \frac{1}{3n}\). Thus, \(\left(1-\frac{1}{3 n}\right)^{n} \approx e^{-\frac{n}{3n}} = e^{-\frac{1}{3}}\).

Step 3: Simplify the Series Terms

The terms \( \left(1-\frac{1}{3 n}\right)^{n} \approx e^{-\frac{1}{3}} \) for large \(n\). So each term of the series approaches a constant \(e^{-\frac{1}{3}}\).

Step 4: Apply the Test for Divergence

If the terms of a series \( a_n \) do not approach 0, i.e., \( \lim_{n \to \infty} a_n eq 0 \), the series diverges. Here, \( \lim_{n \to \infty} \left(1-\frac{1}{3 n}\right)^{n} = e^{-\frac{1}{3}} eq 0\).

Step 5: Conclusion on Convergence

Since the terms approach a non-zero constant \(e^{-\frac{1}{3}}\), the series \( \sum_{n=1}^{\infty}\left(1-\frac{1}{3 n}\right)^{n} \) diverges.

Key Concepts

Test for Divergence

When we want to decide whether an infinite series converges or diverges, one of the first techniques we should consider is the **Test for Divergence**. This test looks at the limit of the sequence of terms that make up the series. If the limit is not zero as the number of terms goes to infinity, the series must diverge.
In simple terms, if we are adding up an infinite number of terms in a series and those terms do not become vanishingly small (tending towards zero), then they can't "pile up" to a finite sum. They instead keep adding up to infinity, causing the series to diverge.
For the series we've been examining, the terms approach a non-zero constant, specifically, they tend towards \(e^{-\frac{1}{3}} \). Since this constant is not zero, the series diverges by the Test for Divergence. This method is critical because it quickly tells us that a series diverges without needing more complex convergence tests.

Exponential Function

Understanding exponential functions is essential for analyzing the given series. At the heart of many complex mathematical phenomena, **Exponential Functions** take the form of \(e^{x}\), where \(e\) is the special mathematical constant approximately equal to 2.71828. This function models growth and decay processes that are multiplicative rather than additive.
In our case, the terms of the series \(\left(1-\frac{1}{3 n}\right)^n \) can be related to the exponential function by manipulating the expression. This manipulation leads us to approximate these terms using the exponential function such that \(\left(1-\frac{1}{3 n}\right)^n \approx e^{-\frac{n}{3n}} = e^{-\frac{1}{3}}\).
This rewrite simplifies analysis since exponential functions have well-understood behavior, including limits and derivatives, making them pivotal in assessing series or sequence convergence.

Limit of a Sequence

The concept of a **Limit of a Sequence** is fundamental in understanding the behavior of the terms in an infinite series. The limit tells us what value the terms approach as the sequence progresses indefinitely. If a sequence has a limit that is zero, it indicates potential for convergence in a series.
For the series at hand, the individual terms are analyzed as \(n\) becomes very large. Here, \(\left(1-\frac{1}{3 n}\right)^{n}\) simplifies to an approximation of \(e^{-\frac{1}{3}}\) for large \(n\). This result shows that the sequence formed by these terms does not approach zero but instead stays around \(e^{-\frac{1}{3}}\).
This constant non-zero limit informs us that the series cannot converge. Without this going to zero, the prospect of adding innately small quantities to reach a finite sum is impossible. Understanding limits in this context is not only about the arithmetic but about recognizing the new behavior introduced by infinity.