Problem 13 - Single Variable Calculus

Question

Determine whether the series converges or diverges. $$ \sum_{n=0}^{\infty} \frac{(-1)^{n+1} e^{n}}{\pi^{n+1}} $$

Step-by-Step Solution

Answer

The given series is an alternating series and can be written as \(\sum_{n=0}^{\infty} \frac{(-1)^{n+1} e^n}{\pi^{n+1}}\). We apply the Alternating Series Test by showing that the sequence \(b_n = \frac{e^n}{\pi^{n+1}}\) is decreasing and that \(\lim_{n \to \infty} b_n = 0\). After showing these conditions are satisfied, we can conclude that the given series converges.

Step 1: Identify the series as alternating

The series can be written as follows: \[ \sum_{n=0}^{\infty} \frac{(-1)^{n+1} e^{n}}{\pi^{n+1}} \] We can see that the series is alternating because the term \((-1)^{n+1}\) changes its sign for every \(n\). Therefore, it is alternating and we can apply the Alternating Series Test.

Step 2: Check the conditions of the Alternating Series Test

To apply the Alternating Series Test, we need to check if the following conditions are both satisfied: 1. \(b_n = \frac{e^{n}}{\pi^{n+1}} \) is a decreasing sequence. 2. \(\lim_{n \to \infty} b_n = 0\) Let's check these conditions.

Step 3: Check if the sequence \(b_n\) is decreasing

We have the sequence \(b_n = \frac{e^{n}}{\pi^{n+1}}\). To show that this is a decreasing sequence, we need to prove that \(b_{n+1} < b_n\), i.e., \( \frac{e^{n+1}}{\pi^{n+2}} < \frac{e^{n}}{\pi^{n+1}}\). Let's simplify this inequality: \[ \frac{e^{n+1}}{\pi^{n+2}} < \frac{e^{n}}{\pi^{n+1}} \\ \Rightarrow \frac{e^{n} e^1}{\pi^{n+1} \pi^1} < \frac{e^{n}}{\pi^{n+1}} \] We can cancel out \(e^{n}\) and \(\pi^{n+1}\): \[ \frac{e}{\pi} < 1 \] Since \(e \approx 2.718\) and \(\pi \approx 3.141\), the inequality holds, which implies that the sequence \(b_n\) is decreasing.

Step 4: Determine the limit of \(b_n\)

Next, we need to find \(\lim_{n \to \infty} b_n = \lim_{n \to \infty}\frac{e^{n}}{\pi^{n+1}}\). To analyze this limit, let's use L'Hôpital's Rule since both the numerator and denominator tend to infinity as n approaches infinity. The derivative of the numerator with respect to \(n\) is \(e^n\), and the derivative of the denominator with respect to \(n\) is \((n+1) \pi^{n} \). Then, applying L'Hôpital's Rule, we get: \[ \lim_{n \to \infty} \frac{e^{n}}{\pi^{n+1}} = \lim_{n \to \infty} \frac{e^{n}}{(n+1) \pi^{n} } \] Now, let's divide both the numerator and denominator by \(e^n\): \[ \lim_{n \to \infty} \frac{1}{(n+1) (\frac{\pi}{e})^{n} } \] Since the exponential term \((\frac{\pi}{e})^{n} \) goes to infinity and the numerator is constant, the limit goes to 0.

Step 5: Apply the Alternating Series Test

As both conditions of the Alternating Series Test are satisfied: 1. \(b_n = \frac{e^{n}}{\pi^{n+1}}\) is a decreasing sequence. 2. \(\lim_{n \to \infty} b_n = 0\) We can conclude that the given series converges.

Key Concepts

Convergence of series

When we talk about the convergence of a series, we mean whether the sum of an infinite sequence of numbers approaches a certain value. For the series given, which is \( \sum_{n=0}^{\infty} \frac{(-1)^{n+1} e^{n}}{\pi^{n+1}} \), convergence means that adding an infinite number of its terms still results in a finite number.

To determine convergence, mathematicians often use tests. Specifically, for alternating series, the **Alternating Series Test** is quite useful. This test provides a straightforward approach to check if such series converges, as it involves two simple conditions:

• The absolute value of the terms must form a decreasing sequence.
• The limit of the terms as \( n \) goes to infinity should be zero.

If these conditions are met, the series can be confidently said to converge. In the case of the series above, both conditions hold true, implying successful convergence. This means that this alternating infinite series adds up to a finite value as terms are continually added.

Decreasing sequence

The concept of a decreasing sequence is essential in determining the convergence of an alternating series. A sequence \( b_n \) is considered decreasing if every term is greater than the term following it. That is, \( b_{n+1} < b_n \) for all \( n \).

In the context of our series, we have \( b_n = \frac{e^{n}}{\pi^{n+1}} \). To show that this is a decreasing sequence, we rewrite the inequality \( \frac{e^{n+1}}{\pi^{n+2}} < \frac{e^{n}}{\pi^{n+1}} \) and simplify:

• Cancel terms to get \( \frac{e}{\pi} < 1 \).
• As \( e \approx 2.718 \) and \( \pi \approx 3.141 \), it's clear that \( e < \pi \), confirming \( \frac{e}{\pi} < 1 \).

This confirms that \( b_n \) forms a decreasing sequence. Such a sequence is crucial because, as per the Alternating Series Test, an alternating series with terms forming a decreasing sequence has the potential to converge.

Limit of a sequence

The limit of a sequence is a foundational concept in understanding series behavior, particularly in convergence proofs. A sequence \( b_n \) has a limit \( L \) if, as \( n \) increases indefinitely, \( b_n \) gets arbitrarily close to \( L \).

For our particular series, we are interested in the limit of the sequence \( b_n = \frac{e^{n}}{\pi^{n+1}} \) as \( n \) approaches infinity. In simpler terms, we want to see what \( \frac{e^{n}}{\pi^{n+1}} \) becomes as \( n \) gets very large.

Using **L'Hôpital's Rule**, a method for evaluating limits of indeterminate forms, we analyze \( \lim_{n \to \infty} \frac{e^{n}}{(n+1) \pi^{n}} \). Upon simplifying, we find:

• The expression reduces to \( \frac{1}{(n+1) (\frac{\pi}{e})^{n}} \).
• As \( n \) increases, \( (\frac{\pi}{e})^{n} \) grows significantly because \( \pi > e \), pushing the fraction's value toward zero.

This result, \( \lim_{n \to \infty} b_n = 0 \), confirms that the sequence tends to zero as \( n \) increases, helping establish the series' convergence through the Alternating Series Test.