Problem 37
Question
Suppose that \(\lim _{n \rightarrow \infty} \sqrt[n]{\left|a_{n}\right|}=L\) and \(L \neq 0\). What is the radius of convergence of the power series \(\sum a_{n} x^{n} ?\)
Step-by-Step Solution
Verified Answer
The radius of convergence (R) of the power series \(\sum a_n x^n\) is given by \(R=L\), where \(L = \lim_{n \rightarrow \infty} \sqrt[n]{|a_n|}\) and \(L \neq 0\).
1Step 1: Recall the definition of radius of convergence
Radius of convergence (R) is defined as the value for which the power series converges. In other words, the power series will converge for all values of \(x\) such that \(|x| < R\) and diverge for all values of \(x\) such that \(|x| > R\).
2Step 2: Apply the Ratio Test
To find the radius of convergence, we will apply the Ratio Test, which tells us that if \(\lim_{n \rightarrow \infty} \left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right| < 1,\) then the power series converges, and if this limit is greater than 1, the power series diverges. So, we need to find the limit:
\[
\lim_{n \rightarrow \infty} \left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right|.
\]
3Step 3: Simplify the limit expression
Now we simplify and rewrite the limit expression:
\[
\lim_{n \rightarrow \infty} \left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right| = \lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right| \cdot \lim_{n \rightarrow \infty} |x|.
\]
4Step 4: Using the given information
We know that \(\lim_{n \rightarrow \infty} \sqrt[n]{|a_n|}=L\) and \(L \neq 0\). From this information, we can find the limit \(\lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right|\) using this formula:
\[
\lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right| = \frac{1}{L}.
\]
5Step 5: Combine the information to find the radius of convergence
With the information we've gathered, we can now find the radius of convergence (R). From Step 3, we have:
\[
\lim_{n \rightarrow \infty} \left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right| = \lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right| \cdot \lim_{n \rightarrow \infty} |x|.
\]
And from Step 4, this simplifies to:
\[
\frac{1}{L} \cdot |x| < 1.
\]
Now, we solve for \(x\) which gives us:
\[
|x| < L.
\]
Thus, the radius of convergence (R) is given by \(R=L.\)
Key Concepts
Power Series ConvergenceRatio TestLimit of a SequenceSequence Convergence
Power Series Convergence
Understanding the convergence of power series is crucial for students embarking on the study of advanced mathematics, particularly in calculus and its applications. A power series is an infinite series in the form of \( \sum a_{n} x^{n} \), where \( a_{n} \) represents the coefficient of the nth term and \( x \) is a variable. Convergence of a power series refers to the series adding up to a finite value when \( x \) is replaced with a particular real number.
For instance,
For instance,
- If \( \sum a_{n} x^{n} \) converges for \( x = c \), then the series sums to a finite number at \( x = c \).
- If \( x \) is within a certain interval, the series converges, and this interval is bounded by the radius of convergence, R. Outside of this interval, the series may diverge, meaning that it doesn't sum to a finite value.
Ratio Test
The Ratio Test is a widely used method for determining the convergence of series, particularly useful for power series. It is applied by comparing the magnitude of successive terms in a series. For a series \( \sum a_{n} \), the Ratio Test uses the limit \( L = \lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right| \).
Here's how you apply it:
Here's how you apply it:
- If \( L < 1 \), the series converges absolutely.
- If \( L > 1 \) (or is infinite), then the series diverges.
- If \( L = 1 \) the test is inconclusive, and convergence cannot be determined by this test alone.
Limit of a Sequence
The limit of a sequence is an important concept in sequences and series, representing the value that the terms of the sequence approach as the index \( n \) increases indefinitely. In mathematical notation, it is expressed as \( \lim_{n \rightarrow \infty} a_n = L \) where \( a_n \) is the nth term of the sequence and \( L \) stands for the limit.
Illustrative examples:
- The sequence \( 1/n \) has a limit of \( 0 \) as \( n \) approaches infinity.
- The sequence \( (1+1/n)^n \) approaches \( e \) (where \( e \) is Euler's number) as \( n \) approaches infinity.
Sequence Convergence
Sequence convergence is the process by which a sequence approaches a particular value, known as its limit, as its index increases. In essence, for a sequence \( a_n \), we say it converges to a value \( L \) if, as \( n \) goes to infinity, the terms \( a_n \) can get as close to \( L \) as we want, provided we go far enough along the sequence.
Characteristics of convergent sequences include:
- Given any small positive number (often called epsilon, \( \epsilon \)), there exists some index \( N \) such that for all \( n > N \), the absolute difference \( |a_n - L| \) is less than \( \epsilon \).
- The terms of the sequence eventually become, and remain, arbitrarily close to the limit \( L \) as \( n \) increases beyond a certain point.
- A sequence that does not converge is said to diverge.
Other exercises in this chapter
Problem 36
Determine whether the sequence \(\left\\{a_{n}\right\\}\) converges or diverges. If it converges, find its limit. \(a_{n}=\frac{n^{p}}{e^{n}}, \quad p>0\)
View solution Problem 37
Determine whether the series is convergent or divergent. \(\sum_{n=1}^{\infty} \frac{n !}{n^{n}}\)
View solution Problem 37
a. Suppose that \(\sum a_{n}\) and \(\sum b_{n}\) are both convergent. Does it follow that \(\Sigma a_{n} b_{n}\) must be convergent? Justify your answer. b. Su
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Euler's Constant a. Show that $$ \ln (n+1) \leq 1+\frac{1}{2}+\cdots+\frac{1}{n} $$ and therefore, \(0
View solution