Problem 60
Question
Convergence or Divergence Which of the series in Exercises \(55-62\) converge, and which diverge? Give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{n^{n}}{\left(2^{n}\right)^{2}}$$
Step-by-Step Solution
Verified Answer
The series diverges because the terms do not approach zero.
1Step 1: Identify the Type of Series
We are given the series \(\sum_{n=1}^{\infty} \frac{n^n}{(2^n)^2}\). This is an infinite series where we need to determine convergence or divergence.
2Step 2: Simplify the General Term
The general term of the series is \(a_n = \frac{n^n}{(2^n)^2} = \frac{n^n}{4^n}\). We can simplify this to \(a_n = (\frac{n}{4})^n\).
3Step 3: Apply the Nth-Term Test
The Nth-term test states that if \(\lim_{{n \to \infty}} a_n eq 0\), then the series diverges. Calculate \(\lim_{{n \to \infty}} (\frac{n}{4})^n\).
4Step 4: Analyze the Limit of \((\frac{n}{4})^n\)
Rewrite \((\frac{n}{4})^n = n^n \cdot (\frac{1}{4})^n\). As \(n \to \infty\), \(n^n\) grows much faster than \(4^n\), implying \((\frac{n}{4})^n \to \infty\). Hence, \(\lim_{{n \to \infty}} (\frac{n}{4})^n eq 0\).
5Step 5: Conclude Divergence
Since \(\lim_{{n \to \infty}} (\frac{n}{4})^n eq 0\), the series diverges according to the Nth-term test.
Key Concepts
Infinite SeriesNth-Term TestLimit AnalysisDivergence
Infinite Series
In the world of mathematics, an infinite series is a sum of infinitely many terms. These terms follow a specific pattern or rule, which we refer to as a sequence. When we talk about the convergence or divergence of a series, we are examining whether the sum of all these infinite terms approaches a specific number, or if it goes on growing indefinitely.
Consider the series \( \sum_{n=1}^{\infty} \frac{n^n}{(2^n)^2} \). Here, each term in the series is given by \( a_n = \frac{n^n}{4^n} \). Infinite series crop up frequently in calculus and other branches of mathematics, both as theoretical constructs and practical tools, like in mathematical models or solving equations.
Consider the series \( \sum_{n=1}^{\infty} \frac{n^n}{(2^n)^2} \). Here, each term in the series is given by \( a_n = \frac{n^n}{4^n} \). Infinite series crop up frequently in calculus and other branches of mathematics, both as theoretical constructs and practical tools, like in mathematical models or solving equations.
Nth-Term Test
The Nth-term test is a simple yet powerful tool for determining the divergence of an infinite series. Here's how it works:
In our problem, the series is \( \sum_{n=1}^\infty (\frac{n}{4})^n \). Applying the Nth-term test, we find that \( \lim_{{n \to \infty}} (\frac{n}{4})^n eq 0 \) since the terms grow infinitely large. Thus, by the Nth-term test, this series diverges.
- If the limit of the nth term \( a_n \) of a series as \( n \) approaches infinity is not zero, that is, \( \lim_{{n \to \infty}} a_n eq 0 \), then the series must diverge.
- However, if the limit equals zero, the test is inconclusive, meaning the series might converge or diverge, and further tests are needed.
In our problem, the series is \( \sum_{n=1}^\infty (\frac{n}{4})^n \). Applying the Nth-term test, we find that \( \lim_{{n \to \infty}} (\frac{n}{4})^n eq 0 \) since the terms grow infinitely large. Thus, by the Nth-term test, this series diverges.
Limit Analysis
Determining the behavior of a series often involves analyzing the limit of its terms, particularly when using the Nth-term test. In this series, we need to calculate the limit of the general term \( a_n = (\frac{n}{4})^n \) as \( n \) approaches infinity.
We first rewrite \((\frac{n}{4})^n\) as \(n^n \cdot (\frac{1}{4})^n\). As \( n \) gets large, \( n^n \) grows extremely fast, much faster than \( 4^n \). This rapid growth means that the terms \((\frac{n}{4})^n\) actually go to infinity. Therefore, the limit is not zero and confirms divergence.
We first rewrite \((\frac{n}{4})^n\) as \(n^n \cdot (\frac{1}{4})^n\). As \( n \) gets large, \( n^n \) grows extremely fast, much faster than \( 4^n \). This rapid growth means that the terms \((\frac{n}{4})^n\) actually go to infinity. Therefore, the limit is not zero and confirms divergence.
- In essence, understanding term growth rates is crucial in limit analysis.
- Identifying which factor grows faster can reveal the nature of the series.
Divergence
Divergence refers to the phenomenon where the terms of a series do not sum to a finite value. For a sequence of terms, when we add them indefinitely and the sum grows without bound, we say the series diverges.
In our series \( \sum_{n=1}^{\infty} (\frac{n}{4})^n \), we've established that it diverges since the terms approach infinity as \( n \) increases. This means no matter how many terms we include, the total sum keeps increasing.
In our series \( \sum_{n=1}^{\infty} (\frac{n}{4})^n \), we've established that it diverges since the terms approach infinity as \( n \) increases. This means no matter how many terms we include, the total sum keeps increasing.
- Recognizing divergence is crucial as it tells us when a series does not settle into a predictable total.
- In mathematical problems, knowing whether a series converges or diverges can help decide the next steps in finding solutions or evaluating problems.
Other exercises in this chapter
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