Chapter 11: Problem 22
Use the limit comparison test to determine whether the series converges or diverges. $$ \sum_{n=1}^{\infty} \frac{2}{3+\sqrt{n}} $$
Short Answer
Expert verified
The series diverges.
Step by step solution
01
Identify the series to compare
The series given is \( \sum_{n=1}^{\infty} \frac{2}{3+\sqrt{n}} \). To perform the limit comparison test, choose a comparable simpler series, like \( \sum_{n=1}^{\infty} \frac{2}{\sqrt{n}} \), since as \( n \) grows large, \( \sqrt{n} \) will dominate the term \( 3+\sqrt{n} \).
02
Establish the comparability
For limit comparison test, define the term \( a_n = \frac{2}{3+\sqrt{n}} \) of the given series and \( b_n = \frac{2}{\sqrt{n}} \), the term of the simple series. The series \( \sum_{n=1}^{\infty} \frac{2}{\sqrt{n}} \) is known to diverge because it's a constant multiple of the harmonic series \( \sum_{n=1}^{\infty} n^{-1/2} \), which diverges.
03
Calculate the limit of the ratio
Compute \( \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{2/(3+\sqrt{n})}{2/\sqrt{n}} \). Simplify this expression to get \( \lim_{n \to \infty} \frac{\sqrt{n}}{3+\sqrt{n}} = \lim_{n \to \infty} \frac{1}{\frac{3}{\sqrt{n}}+1} \).
04
Evaluate the Limit
As \( n \to \infty \), \( \frac{3}{\sqrt{n}} \) approaches 0. Thus, the limit becomes \( \lim_{n \to \infty} \frac{1}{1+0} = 1 \). The limit is finite and non-zero.
05
Conclude with the Limit Comparison Test
According to the limit comparison test, if \( \lim_{n \to \infty} \frac{a_n}{b_n} = c \) where \( 0 < c < \infty \), both series \( \sum_{n=1}^{\infty} a_n \) and \( \sum_{n=1}^{\infty} b_n \) either converge or diverge together. Since \( \sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} \frac{2}{\sqrt{n}} \) diverges, the original series \( \sum_{n=1}^{\infty} \frac{2}{3+\sqrt{n}} \) also diverges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Series Convergence
In the world of infinite series, understanding whether a series converges or diverges is crucial. A series converges when the sum of its infinite terms approaches a specific value. For example, the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges, because the sums of its terms get closer and closer to a finite number.
Convergent series have several important properties:
Convergent series have several important properties:
- The terms get smaller and approach zero as the series progresses.
- The partial sums of the series stabilize and reach a particular value.
Divergent Series
A divergent series behaves quite differently from a convergent one. Divergence means that as you add more terms, the sum doesn’t settle to any number — it might grow to infinity or oscillate without settling. Take the harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \), for example. This series diverges because its partial sums increase indefinitely.
Divergent series are characterized by:
Divergent series are characterized by:
- The partial sums do not approach a fixed number.
- They often have terms that don't shrink down to zero fast enough.
Harmonic Series
A harmonic series is a fascinating yet simple series defined by \( \sum_{n=1}^{\infty} \frac{1}{n} \). Though it might seem that its terms shrink to zero, which could suggest convergence, this series is actually a classic example of divergence. It's named for its connection to musical harmony and is used as a benchmark in various tests for series convergence.
The harmonic series can be modified or extended:
The harmonic series can be modified or extended:
- Substitute \( n^{-1} \) with \( n^{-p} \) for positive \( p \).
- If \( p > 1 \), the series converges but if \( 0 < p \leq 1 \), the series diverges.