Problem 35 - Essential Calculus Early Transcendentals

Question

\(19-40=\) Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)^{n^{2}} $$

Step-by-Step Solution

Answer

The series diverges.

Step 1: Understanding the terms

The series given is \( \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)^{n^{2}} \). Each term of the series is \( a_n = \left(1+\frac{1}{n}\right)^{n^{2}} \). Our task is to determine if the series is absolutely convergent, conditionally convergent, or divergent.

Step 2: Analyzing the behavior

Observe that as \( n \to \infty \), \( \left(1+\frac{1}{n}\right)^{n} \to e \), but our terms raise this expression to the power of \( n^2 \). Thus, \( a_n = \left(1+\frac{1}{n}\right)^{n^{2}} \to \infty \) as \( n \to \infty \).

Step 3: Testing for divergence

Because \( a_n \to \infty \), the terms do not approach zero as \( n \to \infty \). According to the divergence test, if the limit of \( a_n \) is not zero, the series \( \sum_{n=1}^{\infty} a_n \) must diverge.

Key Concepts

Divergence Test

The divergence test is a fundamental tool in determining whether a series converges or diverges. It's helpful to first understand that convergence in series means that adding up an infinite number of terms results in a finite number. A simple way to initially test for divergence is by examining the limit of the sequence's terms, denoted as \( a_n \).

In the given exercise, the terms are \( a_n = \left(1 + \frac{1}{n} \right)^{n^2} \). By investigating this, we see that as \( n \to \infty \), the expression \( \left(1 + \frac{1}{n} \right)^n \to e \) (where \( e \) is Euler's number, approximately 2.718). When raised to a power that grows as \( n^2 \), like in our series, it grows very quickly and does not approach zero.

**Key Concepts:**- If the limit \( \lim_{n \to \infty} a_n eq 0 \), the series \( \sum a_n \) diverges.- In this case, since \( a_n \to \infty \), the series does not converge to a finite number but diverges. Thus, the divergence test confirms that our series is divergent.

Absolutely Convergent

A series \( \sum a_n \) is said to be absolutely convergent if the series of absolute values \( \sum |a_n| \) converges. Absolute convergence is a stronger form of convergence, implying that no matter how you arrange the terms, the series will still converge.

For our series, absolute convergence isn't applicable because of what we found using the divergence test. The term \( \left(1 + \frac{1}{n} \right)^{n^2} \) grows without bound as \( n \to \infty \), meaning that \( |a_n| \) would also diverge.

Thus, the notion of absolute convergence doesn't apply here because the series doesn't satisfy the primary condition - the infinite series of its absolute terms should sum up to a finite value. Since the divergent nature of the term \( a_n \) has already been established, absolute convergence is ruled out.

Conditionally Convergent

A series is conditionally convergent if it converges, but its absolute value series \( \sum |a_n| \) does not converge. This is usually seen in alternating series where the terms get smaller in absolute value but switch signs.

Given that we've established through the divergence test that the series of \( \left(1 + \frac{1}{n} \right)^{n^2} \) does not converge, there's no possibility for conditional convergence because the first step for any kind of convergence is already unmet.

To identify a conditionally convergent series, one would typically look for the presence of changing signs and a diminishing sequence in the absolute values of the terms, neither of which applies in this case. Hence, conditional convergence is not a characteristic of our given series.