Problem 61 - Thomas Calculus

Question

Which of the series in Exercises \(57-64\) converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{n^{n}}{2^{\left(n^{2}\right)}} $$

Step-by-Step Solution

Answer

The series converges by the root test.

Step 1: Identify the Series

The given series is \( \sum_{n=1}^{\infty} \frac{n^{n}}{2^{(n^2)}} \). We need to determine whether this series converges or diverges.

Step 2: Apply the Root Test

The root test is often useful when terms involve powers in their expressions. The root test states that for a series \( \sum a_n \), if \( \lim_{n \to \infty} \sqrt[n]{|a_n|} = L \), then the series converges if \( L < 1 \) and diverges if \( L > 1 \). If \( L = 1 \), the test is inconclusive.

Step 3: Analyze \(a_n = \frac{n^n}{2^{n^2}}\) for the Root Test

For \(a_n = \frac{n^n}{2^{n^2}}\), we compute \( \sqrt[n]{a_n} = \left(\frac{n^n}{2^{n^2}}\right)^{1/n} = \frac{n}{2^{n}}^{1/n} \). Simplifying, we get \( \sqrt[n]{a_n} = \frac{n^{1/n}}{2^n} \).

Step 4: Compute the Limit

Now we compute the limit: \[ \lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \frac{n^{1/n}}{2^n}\]As \( n \to \infty \), the term \( n^{1/n} \to 1 \) and \( 2^n \to \infty \), hence \[ \lim_{n \to \infty} \frac{n^{1/n}}{2^n} = 0\]

Step 5: Conclude Using the Root Test

Since \( \lim_{n \to \infty} \sqrt[n]{a_n} = 0 \), which is less than 1, the series \( \sum_{n=1}^{\infty} \frac{n^n}{2^{n^2}} \) converges by the root test.

Key Concepts

Root Test

The root test is a powerful tool used to determine the convergence or divergence of infinite series. It's particularly handy when dealing with series where each term involves powers. The test considers a series of the form \( \sum a_n \), and the critical factor to look for here is the limit \( \lim_{n \to \infty} \sqrt[n]{|a_n|} = L \). The rule of thumb is quite straightforward:

• If \( L < 1 \), the series converges.
• If \( L > 1 \), the series diverges.
• If \( L = 1 \), the test is inconclusive, and further analysis is needed.

In our specific example with series \( \sum_{n=1}^{\infty} \frac{n^n}{2^{n^2}} \), we applied the root test. Calculating the nth root of \( a_n \), followed by the limit as \( n \to \infty \), showed us that this series converges since the limit approached 0, which is less than 1.

Infinite Series

An infinite series is essentially the sum of infinitely many numbers written in sequence. Formally, we use the notation \( \sum_{n=1}^{\infty} a_n \) to represent the infinite series with terms \( a_n \). As the sequence progresses, the concept of convergence or divergence comes into play.

• Convergence: If adding the terms sequentially leads to a finite number, the series converges.
• Divergence: Conversely, if the sum doesn't settle on a finite value, the series is said to diverge.

For the series \( \sum_{n=1}^{\infty} \frac{n^n}{2^{n^2}} \), we concluded it converges based on the root test. The behavior of each term shrinks significantly as \( n \) becomes large, driven by the rapid growth of the denominator, \( 2^{n^2} \).

Limit

The concept of a limit is integral to calculus and analysis. Limits help us understand the behavior of functions as inputs approach some value. Specifically for series, limits are crucial in determining convergence.Consider a sequence derived from the root test: \( \lim_{n \to \infty} \sqrt[n]{a_n} = L \). We study the behavior of \( \sqrt[n]{a_n} \) as \( n \) increases. If the limit \( L \) is known, it tells us how the series behaves. We apply limits directly in the root test:

• Using \( a_n = \frac{n^n}{2^{n^2}} \), the sequence \( \sqrt[n]{a_n} \) simplifies to \( \frac{n^{1/n}}{2^n} \).
• The limit turns to \( 0 \) as \( n \) approaches infinity, indicating convergence.

Acknowledging limits gives clarity on series' potential to converge or diverge, vital to rigorous mathematical conclusions.

Power Series

Power series are special types of infinite series where each term is a function of a variable raised to integer powers. They're typically represented as \( \sum_{n=0}^{\infty} c_n x^n \), where \( c_n \) are coefficients and \( x \) can vary.These series are highly useful in approximating functions and are prominent in mathematical calculus:

• They expand into polynomials of infinite degree, allowing smooth function approximations over intervals.
• Convergence is heavily dependent on the value of the variable \( x \) and the coefficients \( c_n \).

Though the specific exercise \( \sum_{n=1}^{\infty} \frac{n^n}{2^{n^2}} \) isn't a power series, understanding them gives a broader context of the types of series and methodologies used in the analysis, underpinning concepts like the root test and investigations into patterns of convergence.