Problem 15
Question
\(9-32\) n Determine whether the sequence converges or diverges. If it converges, find the limit. $$a_{n}=\frac{n^{2}}{\sqrt{n^{3}+4 n}}$$
Step-by-Step Solution
Verified Answer
The sequence diverges.
1Step 1: Analyze the Sequence
The given sequence is \(a_{n} = \frac{n^{2}}{\sqrt{n^{3}+4n}}\). To determine whether this sequence converges or diverges, we need to evaluate the limit of \(a_{n}\) as \(n\) approaches infinity.
2Step 2: Simplify the Expression
To simplify \(\frac{n^{2}}{\sqrt{n^{3}+4n}}\), factor \(n^{3}\) out of the square root in the denominator: \(\sqrt{n^{3}+4n} = \sqrt{n^{3}(1 + \frac{4}{n^{2}})} = n^{3/2}\sqrt{1 + \frac{4}{n^{2}}}\). This allows us to rewrite the sequence as: \(a_{n} = \frac{n^{2}}{n^{3/2}\sqrt{1 + \frac{4}{n^{2}}}} = \frac{n^{1/2}}{\sqrt{1 + \frac{4}{n^{2}}}}\).
3Step 3: Evaluate Limit as n Approaches Infinity
Consider the behavior of the term \(\sqrt{1 + \frac{4}{n^{2}}}\) as \(n\) approaches infinity. As \(n\) becomes very large, \(\frac{4}{n^{2}}\) approaches zero, leading to \(\sqrt{1 + \frac{4}{n^{2}}} \to 1\). Thus, \(a_{n}\) can be approximated by \(\frac{n^{1/2}}{1} = n^{1/2}\).
4Step 4: Determine Convergence or Divergence
Since \(n^{1/2}\) approaches infinity as \(n\) becomes very large, the sequence \(a_{n} = \frac{n^{2}}{\sqrt{n^{3}+4n}}\) diverges.
Key Concepts
Limit EvaluationSimplifying ExpressionsDivergence of Sequences
Limit Evaluation
When assessing sequence convergence, one crucial step is to evaluate the limit as the sequence parameters approach infinity. For the given sequence \( a_{n} = \frac{n^{2}}{\sqrt{n^{3}+4n}} \), we need to determine the behavior as \( n \rightarrow \infty \).
To start, consider dominant terms both in the numerator and the denominator, as these primarily dictate the behavior at infinity. The numerator here is \( n^2 \) and in the denominator, \( \sqrt{n^{3}} \) dominates since it grows faster than \( \sqrt{4n} \).
Ultimately, as \( n \) becomes exceedingly large, less significant terms become negligible, simplifying the evaluation of limits. This is why we focus on \( \frac{n^2}{n^{3/2}\sqrt{1+\frac{4}{n^{2}}}} \). Recognizing that \( \sqrt{1+\frac{4}{n^{2}}} \rightarrow 1 \) assists in identifying that the limit involves \( \frac{n^{1/2}}{1} = n^{1/2} \), clarifying that the sequence doesn't converge but rather diverges as \( n \rightarrow \infty \).
To start, consider dominant terms both in the numerator and the denominator, as these primarily dictate the behavior at infinity. The numerator here is \( n^2 \) and in the denominator, \( \sqrt{n^{3}} \) dominates since it grows faster than \( \sqrt{4n} \).
Ultimately, as \( n \) becomes exceedingly large, less significant terms become negligible, simplifying the evaluation of limits. This is why we focus on \( \frac{n^2}{n^{3/2}\sqrt{1+\frac{4}{n^{2}}}} \). Recognizing that \( \sqrt{1+\frac{4}{n^{2}}} \rightarrow 1 \) assists in identifying that the limit involves \( \frac{n^{1/2}}{1} = n^{1/2} \), clarifying that the sequence doesn't converge but rather diverges as \( n \rightarrow \infty \).
Simplifying Expressions
Simplifying mathematical expressions, especially with complex terms, is crucial for understanding sequence convergence.
In the expression \( a_{n} = \frac{n^{2}}{\sqrt{n^{3}+4n}} \), simplifying requires strategic factoring. By factoring out \( n^3 \) within the square root, we transform it to \( \sqrt{n^3(1+\frac{4}{n^2})} \), simplifying our original expression.
This strategic rearrangement helps expose the term that majorly affects growth: \( n^{3/2} \). Now, the sequence is rewritten as:
As \( n \) increases, this further simplifies the limit evaluation since \( \sqrt{1+\frac{4}{n^2}} \) effectively becomes \( 1 \), aiding in clearer determination of convergence or divergence of the sequence.
In the expression \( a_{n} = \frac{n^{2}}{\sqrt{n^{3}+4n}} \), simplifying requires strategic factoring. By factoring out \( n^3 \) within the square root, we transform it to \( \sqrt{n^3(1+\frac{4}{n^2})} \), simplifying our original expression.
This strategic rearrangement helps expose the term that majorly affects growth: \( n^{3/2} \). Now, the sequence is rewritten as:
- Numerator: \( n^2 \)
- Denominator: \( n^{3/2}\sqrt{1+\frac{4}{n^2}} \)
As \( n \) increases, this further simplifies the limit evaluation since \( \sqrt{1+\frac{4}{n^2}} \) effectively becomes \( 1 \), aiding in clearer determination of convergence or divergence of the sequence.
Divergence of Sequences
Divergence refers to sequences that do not settle into a steady state as \( n \rightarrow \infty \). For sequences like \( a_{n}=\frac{n^{2}}{\sqrt{n^{3}+4n}} \), showing they diverge involves demonstrating that they do not approach a finite limit.
When analyzing, if the resulting simplified expression grows without bounds, such as \( n^{1/2} \), we infer divergence. This means its terms grow indefinitely instead of clustering around a particular value.
With \( n^{1/2} \), it clearly shows that the terms get infinitely large as \( n \) increases—a classic sign of divergence. When solving such problems, always look for their growth tendency as \( n \) heads to infinity. This helps determine if they diverge or converge.
Remember, convergence means tending towards a certain number, while divergence indicates growing infinitely or oscillating without settling.
When analyzing, if the resulting simplified expression grows without bounds, such as \( n^{1/2} \), we infer divergence. This means its terms grow indefinitely instead of clustering around a particular value.
With \( n^{1/2} \), it clearly shows that the terms get infinitely large as \( n \) increases—a classic sign of divergence. When solving such problems, always look for their growth tendency as \( n \) heads to infinity. This helps determine if they diverge or converge.
Remember, convergence means tending towards a certain number, while divergence indicates growing infinitely or oscillating without settling.
Other exercises in this chapter
Problem 15
\(13-16=\) Approximate the sum of the series correct to four decimal places. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1} n^{2}}{10^{n}} $$
View solution Problem 15
Find the radius of convergence and interval of convergence of the series. $$\sum_{n=0}^{\infty} \frac{(x-2)^{n}}{n^{2}+1}$$
View solution Problem 16
Find a power series representation for the function and determine the radius of convergence. $$ f(x)=x^{2} \tan ^{-1}\left(x^{3}\right) $$
View solution Problem 16
Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$\sum_{n=1}^{\infty} \frac{1+3^{n}}{2^{n}}$$
View solution