Chapter 9: Problem 36
Converge, and which diverge? Use any method, and give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{n+2^{n}}{n^{2} 2^{n}}$$
Short Answer
Expert verified
The series converges.
Step by step solution
01
Simplify the expression
The given series is \( \sum_{n=1}^{\infty} \frac{n + 2^n}{n^2 2^n} \). We can simplify the term: \( \frac{n}{n^2 2^n} + \frac{2^n}{n^2 2^n} = \frac{n}{n^2 2^n} + \frac{1}{n^2} \). This simplifies to \( \sum_{n=1}^{\infty} \left( \frac{1}{n 2^n} + \frac{1}{n^2} \right) \).
02
Consider the series individually
This results in two separate series: \( \sum_{n=1}^{\infty} \frac{1}{n 2^n} \) and \( \sum_{n=1}^{\infty} \frac{1}{n^2} \). Each can be considered individually for convergence or divergence.
03
Assess the second series
The series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is a well-known convergent p-series, where \( p = 2 > 1 \). Therefore, \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges.
04
Assess the first series - Ratio test
For the series \( \sum_{n=1}^{\infty} \frac{1}{n 2^n} \), apply the ratio test. Consider the ratio of consecutive terms: \( \lim_{n \to \infty} \left| \frac{\frac{1}{(n+1) 2^{n+1}}}{\frac{1}{n 2^n}} \right| = \lim_{n \to \infty} \frac{n}{n+1} \cdot \frac{1}{2} = \frac{1}{2} \). Since this limit is less than 1, the series \( \sum_{n=1}^{\infty} \frac{1}{n 2^n} \) converges.
05
Combine results for overall convergence
Both series \( \sum_{n=1}^{\infty} \frac{1}{n 2^n} \) and \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converge. Therefore, the original series \( \sum_{n=1}^{\infty} \left( \frac{1}{n 2^n} + \frac{1}{n^2} \right) \) converges as the sum of convergent series.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ratio Test
The Ratio Test is a useful analytical tool in determining the convergence of infinite series. It examines the ratio of consecutive terms in a sequence to ascertain if the series converges or diverges.
The Ratio Test involves the following steps:
The Ratio Test involves the following steps:
- Take the absolute value of the ratio of consecutive terms. If \(m_k\) is the \(k\)-th term of the series, consider \(\lim_{k \to \infty} \left| \frac{m_{k+1}}{m_k} \right|\).
- If this limit is less than 1, the series converges absolutely.
- If the limit is greater than 1, the series diverges.
- If the limit equals 1, the test is inconclusive.
p-series
A p-series is a specific form of infinite series that takes the expression \(\sum_{n=1}^{\infty} \frac{1}{n^p}\). The convergence of a p-series critically depends on the value of \(p\). Here are the essential points to understand:
- If \(p > 1\), the p-series converges.
- If \(p \leq 1\), the p-series diverges.
Convergence and Divergence Analysis
Convergence and divergence are crucial concepts to understanding the behavior of infinite series. Essentially, a convergent series approaches a finite limit, while a divergent series does not.
To determine whether a series converges or diverges:
This leads to the final takeaway: if individual series converge, their sum also converges. This process of convergence and divergence analysis reveals the complex nature of series while providing structured methods to tackle them.
To determine whether a series converges or diverges:
- Break down the series into simpler components, if possible – this can make analysis more manageable.
- Use tests like the Ratio Test or apply known convergent series like p-series for comparison.
- Sometimes a combination of tests may be needed to conclude correctly.
This leads to the final takeaway: if individual series converge, their sum also converges. This process of convergence and divergence analysis reveals the complex nature of series while providing structured methods to tackle them.