Problem 12
Question
Test the series for convergence or divergence. \( \displaystyle \sum_{k = 1}^{\infty} \frac {1}{k \sqrt{k^2 +1}} \)
Step-by-Step Solution
Verified Answer
The series \( \sum_{k=1}^{\infty} \frac{1}{k \sqrt{k^2+1}} \) converges.
1Step 1: Choose a Test for Convergence
To determine the convergence of the series, we can use the Limit Comparison Test, as it involves a series of terms with a rational expression, which is typical for such tests.
2Step 2: Compare with a Known Series
Consider the series \( \sum_{k=1}^{\infty} \frac{1}{k^2} \), which is a known convergent p-series with \( p = 2 \). We will compare our series to this using the limit comparison test.
3Step 3: Apply the Limit Comparison Test
Calculate the limit of the ratio of the given series term to the term of the comparison series:\[L = \lim_{k \to \infty} \frac{1 / (k \sqrt{k^2 + 1})}{1 / k^2} = \lim_{k \to \infty} \frac{k^2}{k \sqrt{k^2 + 1}}\]Simplifying this expression gives:\[L = \lim_{k \to \infty} \frac{k}{\sqrt{k^2 + 1}} = \lim_{k \to \infty} \frac{1}{\sqrt{1 + \frac{1}{k^2}}}\]As \(k\) approaches infinity, \(\frac{1}{k^2}\) approaches 0, so:\[L = \frac{1}{\sqrt{1 + 0}} = 1\]
4Step 4: Conclude with the Limit Comparison Test
Since the limit \(L = 1\) is a finite positive number, and \(\sum_{k=1}^{\infty} \frac{1}{k^2}\) is a convergent series, by the Limit Comparison Test, the original series \( \sum_{k = 1}^{\infty} \frac {1}{k \sqrt{k^2 +1}} \) also converges.
Key Concepts
Limit Comparison TestP-SeriesSeries ConvergenceInfinite Series
Limit Comparison Test
The Limit Comparison Test is a method used to determine the convergence or divergence of an infinite series by comparing it to another series that is already known to converge or diverge. This test is particularly useful when dealing with series that have complicated terms. The idea is to:
- Select a known series which resembles the complicated series.
- Calculate the limit of the ratio of the terms from both series as the index approaches infinity.
P-Series
A p-series is a type of infinite series that has the form:\[ \sum_{k=1}^{\infty} \frac{1}{k^p} \]where \( p \) is a positive real number. The convergence of a p-series depends on the value of \( p \):
- The series converges if \( p > 1 \).
- The series diverges if \( p \leq 1 \).
Series Convergence
Series convergence refers to whether the sum of an infinite series converges to a finite limit or not. Convergence tests are the various methods used to determine the behavior of these series. It's essential for mathematical analysis because:
- If a series converges, its terms approach 0 as the index tends to infinity.
- If a series diverges, the sum grows unbounded or does not settle to a single value.
Infinite Series
An infinite series is the sum of an infinite sequence of terms. Infinite series can be expressed in the form:\[ a_1 + a_2 + a_3 + \cdots = \sum_{k=1}^{\infty} a_k \]Infinite series are generally classified into different types, including geometric, arithmetic, harmonic, and p-series. Understanding infinite series is fundamental because it allows us to:
- Model natural phenomena and apply them in scientific computations.
- Explore different approximation techniques and numerical analysis methods.
- Make predictions and verify asymptotic properties of functions.
Other exercises in this chapter
Problem 12
Express the function as the sum of a power series by first using partial fractions. Find the interval of convergence. \( f(x) = \frac {2x + 3}{x^2 + 3x + 2} \)
View solution Problem 12
Find the radius of convergence and interval of convergence of the series. \( \sum_{n = 1}^{\infty} \frac {( - 1)^{n-1}}{n5^n} x^n \)
View solution Problem 12
Use the Ratio Test to determine whether the series is convergent or divergent. \( \displaystyle \sum_{k = 1}^{\infty} ke^{- k} \)
View solution Problem 12
Determine whether the series converges or diverges. \( \displaystyle \sum_{k = 1}^{\infty} \frac {(2k - 1)(k^2 - 1)}{(k + 1)(k^2 + 4)^2} \)
View solution