Problem 59 - Thomas Calculus

Question

Convergence or Divergence Which of the series in Exercises \(55-62\) 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 ratio test.

Step 1: Identify the Series

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

Step 2: Analyze the General Term

The general term of the series is \( a_n = \frac{n^{n}}{2^{(n^2)}} \). We will analyze \( a_n \) to understand its behavior as \( n \) approaches infinity.

Step 3: Apply the Ratio Test

For the ratio test, calculate \( \frac{a_{n+1}}{a_n} \):\[\frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)^{n+1}}{2^{((n+1)^2)}}}{\frac{n^n}{2^{n^2}}} = \frac{(n+1)^{n+1}}{n^n} \cdot \frac{2^{n^2}}{2^{((n+1)^2)}}.\] Simplifying this expression gives: \[\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{n^n} \cdot \frac{1}{2^{2n+1}}.\]

Step 4: Simplify the Limit Expression

Further simplify and evaluate the limit as \( n \) approaches infinity: \[\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{(n+1)^{n+1}}{n^n} \cdot \frac{1}{2^{2n+1}} = \lim_{n \to \infty}\left(\left(1 + \frac{1}{n}\right)^n (n+1) \cdot \frac{1}{2^{2n+1}}\right).\]

Step 5: Evaluate the Limit

Using the approximation \( \left(1 + \frac{1}{n}\right)^n \) approaches \( e \) as \( n \to \infty \), we rewrite the expression: \[\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{e(n+1)}{2^{2n+1}} = 0.\] Since the ratio test gives a limit of 0, which is less than 1, the series converges.

Key Concepts

Ratio Test

The Ratio Test is a popular method used to determine the convergence or divergence of an infinite series. It examines the limit of the ratio of the terms in the series as the index approaches infinity.
Here's how it works:

• Take the general term of the series, denoted as \( a_n \).
• Find the next term, \( a_{n+1} \).
• Calculate the ratio \( \frac{a_{n+1}}{a_n} \).
• Determine the limit of this ratio as \( n \) approaches infinity: \( \lim_{n \to \infty} \frac{a_{n+1}}{a_n} \).

The key here is the limit interpretation:

• If the limit is less than 1, the series converges.
• If the limit is greater than 1, or is infinite, the series diverges.
• If the limit equals 1, the test is inconclusive.

In our exercise, the limit of the ratio was zero, a clear indication of convergence because it is less than 1.

Infinite Series

Infinite series are sums that continue indefinitely. Understanding these can seem tricky because they don't have a finite number of terms.
To grasp them better, consider these basics:

• A series is essentially an expression that adds terms sequentially, such as \( a_1 + a_2 + a_3 + \ldots \).
• An infinite series does not stop, meaning it has an endless number of terms.

Take the series from our problem, \( \sum_{n=1}^{\infty} \frac{n^n}{2^{n^2}} \). It's an infinite series due to the infinite sum sign \( \sum_{n=1}^{\infty} \), indicating we add terms like \( \frac{1}{2}, \frac{4}{16}, \frac{27}{512}, \ldots \) forever.
Infinite series can converge (approach a specific value) or diverge (grow without bound or oscillate) depending on the nature of their terms and summation.

Series Convergence

Series convergence is a central concept in understanding infinite series. It determines whether the sum of all terms in the series reaches a finite limit or not.

Here's how we can check for convergence:

• A series converges if the sequence of its partial sums approaches a finite number as the number of terms goes to infinity.
• Methods such as the Ratio Test help in checking convergence by simplifying the approach.

For example, in our series \( \sum_{n=1}^{\infty} \frac{n^n}{2^{n^2}} \), evaluating it with the Ratio Test revealed that the limit was less than 1, confirming its convergence.
Convergence implies we can meaningfully discuss the sum of the infinite series, whereas divergence suggests the series lacks a finite sum.