Chapter 11: Problem 16
Use a basic comparison test to determine whether the series converges or diverges. $$ \sum_{n=1}^{\infty} \frac{2+\cos n}{n^{2}} $$
Short Answer
Expert verified
The series converges by the comparison test.
Step by step solution
01
Compare Original Series to Known Series
First, identify a simpler series to compare with. Choose \( \sum_{n=1}^{\infty} \frac{1}{n^{2}} \) because it is known to converge by the p-series test with \( p = 2 > 1 \).
02
Establish the Inequality
For each term of the original series \( \frac{2 + \cos n}{n^2} \), notice that \(-1 \leq \cos n \leq 1\). Therefore, \( 1 \leq 2 + \cos n \leq 3 \). This gives \( \frac{1}{n^2} \leq \frac{2+\cos n}{n^2} \leq \frac{3}{n^2} \).
03
Apply the Comparison Test
We have shown that \( \frac{1}{n^2} \leq \frac{2+\cos n}{n^2} \leq \frac{3}{n^2} \). The series \( \sum_{n=1}^{\infty} \frac{1}{n^{2}} \) converges and so does \( \sum_{n=1}^{\infty} \frac{3}{n^{2}} \), by the p-series test. Hence, by the comparison test, the original series \( \sum_{n=1}^{\infty} \frac{2 + \cos n}{n^2} \) converges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Comparison Test
The comparison test is a handy tool used to determine the convergence or divergence of a series. It involves comparing a complex series with a simpler, known series. If a series resembles a known convergent series closely enough, it is likely also convergent. Similarly, if it resembles a divergent series, it might diverge too.
The idea is simple:
The idea is simple:
- If you can find a larger series that is convergent, the series you are examining is also convergent.
- If you find a smaller series that diverges, the series you are testing also diverges.
P-Series
A p-series is a specific type of infinite series expressed as \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). These series are particularly notable because they provide clear convergence criteria based on the exponent \( p \).
Key characteristics include:
Key characteristics include:
- When \( p > 1 \), the series converges. This is because, as n grows larger, the terms of the series shrink quickly enough towards zero.
- When \( p \leq 1 \), the series diverges because terms shrink too slowly.
Cosine Function
The cosine function plays a critical part in forming the original series terms in the exercise example. Cosine, being a periodic trigonometric function, oscillates between -1 and 1. Its periodic nature and strict bounds are useful traits when attempting inequality comparisons.
Understanding these properties:
Understanding these properties:
- Since \(-1 \leq \cos n \leq 1\), adding 2 results in \(1 \leq 2 + \cos n \leq 3\).
- This reformation keeps the trigonometric function in check within a defined range, allowing for reliable comparison with typical p-series expressions.
Mathematical Inequality
Mathematical inequalities are indispensable tools in determining series convergence through comparison. In this context, inequalities help establish a relation between the original series and a benchmark series.
Why they matter:
Why they matter:
- In the series \( \frac{2+\cos n}{n^2} \), the inequality \( 1 \leq 2 + \cos n \leq 3 \) translates into term-wise inequalities like \( \frac{1}{n^2} \leq \frac{2+\cos n}{n^2} \leq \frac{3}{n^2} \).
- This affirms that as the bounding series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) and \( \sum_{n=1}^{\infty} \frac{3}{n^2} \) converge, the original series \( \sum_{n=1}^{\infty} \frac{2+\cos n}{n^2} \) must converge too.