Problem 12
Question
For \(k \in \mathbb{N}\), let \(a_{2 k-1}:=4^{k-1} / 9^{k-1}\) and \(a_{2 k}:=4^{k-1} / 9^{k} .\) Show that \(\left|a_{2 k} / a_{2 k-1}\right|=\frac{1}{9}\) and \(\left|a_{2 k+1} / a_{2 k}\right|=4\) for all \(k \in \mathbb{N}\) and so the Ratio Test for the convergence of \(\sum_{k=1}^{\infty} a_{k}\) is inconclusive. Prove that \(\left|a_{k}\right|^{1 / k} \rightarrow \frac{2}{3}\) as \(k \rightarrow \infty\) and use the Root Test to conclude that \(\sum_{k=1}^{\infty} a_{k}\) is convergent.
Step-by-Step Solution
Verified Answer
We found the absolute values of the ratios of consecutive terms and showed that \(\left|a_{2k}/a_{2k-1}\right| = \frac{1}{9}\) and \(\left|a_{2k+1}/a_{2k}\right| = 4\) for all \(k \in \mathbb{N}\). Since these values aren't constant, the Ratio Test is inconclusive. Computing the limit for the Root Test, we found that \(\lim_{k \rightarrow \infty} \left|a_k\right|^{1/k} = \frac{2}{3}\), which is less than 1. Therefore, by the Root Test, the series \(\sum_{k=1}^\infty a_k\) converges.
1Step 1: Find the ratios of consecutive terms for both sequences
To find the Ratio Test ratios, we want to compute \(\left|a_{2k}/a_{2k-1}\right|\) and \(\left|a_{2k+1}/a_{2k}\right|\) for the given sequences.
For \(\left|a_{2k}/a_{2k-1}\right|\), we have:
\(\left|\frac{4^{k-1}/9^k}{4^{k-1}/9^{k-1}}\right| = \left|\frac{4^{k-1}}{9^{k}}\right|\times \left|\frac{9^{k-1}}{4^{k-1}}\right| = |9^{-1}| = \frac{1}{9}\)
For \(\left|a_{2k+1}/a_{2k}\right|\), we have:
\(\left|\frac{4^k/9^k}{4^{k-1}/9^k}\right| = \left|\frac{4^k}{4^{k-1}}\right| = 4\)
Thus, we have shown that \(\left|a_{2k}/a_{2k-1}\right| = \frac{1}{9}\) and \(\left|a_{2k+1}/a_{2k}\right| = 4\) for all \(k \in \mathbb{N}\).
2Step 2: The Ratio Test is inconclusive
Since the absolute values of the ratios of consecutive terms aren't constant, the Ratio Test cannot determine the convergence of the series.
3Step 3: Compute the limit for the Root Test
To apply the Root Test, we need to find the limit of \(\left|a_k\right|^{1/k}\) as \(k \rightarrow \infty\). Since the series has different expressions for the even and odd terms, we can compute the separate limits for both even and odd k.
For odd k, \(k = 2n - 1\) for some \(n \in \mathbb{N}\):
\(\lim_{n\to\infty}\left(\frac{4^{n-1}}{9^{n-1}}\right)^{1/(2n-1)} = \lim_{n\to\infty}\left(\frac{2^{2(n-1)}}{3^{2(n-1)}}\right)^{1/(2n-1)} = \lim_{n\to\infty}\left(\frac{2}{3}\right)^{2/(2n-1)} = \left(\frac{2}{3}\right)^2\)
For even k, \(k = 2n\) for some \(n \in \mathbb{N}\):
\(\lim_{n\to\infty}\left(\frac{4^{n-1}}{9^n}\right)^{1/(2n)} = \lim_{n\to\infty}\left(\frac{2^{2(n-1)}}{3^{2n}}\right)^{1/(2n)} = \lim_{n\to\infty}\left(\frac{2}{3}\right)^{2n/(2n)} = \frac{2}{3}\)
Since both limits are equal, we conclude that \(\lim_{k\to\infty} \left|a_k\right|^{1/k} = \frac{2}{3}\).
4Step 4: Apply the Root Test
We have shown that \(\lim_{k\to\infty} \left|a_k\right|^{1/k} = \frac{2}{3}\), which is less than 1. By the Root Test, series \(\sum_{k=1}^\infty a_k\) converges.
Key Concepts
Ratio TestRoot TestSequences and SeriesConvergence of Series
Ratio Test
The Ratio Test is a popular method in calculus used to determine whether a series converges or diverges. You calculate the limit of the absolute value of the ratio of consecutive terms. Specifically, for a series \( \sum a_k \), you find \( \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| \).
This can tell us:
This can tell us:
- If the limit is less than 1, the series converges absolutely.
- If the limit is greater than 1, the series diverges.
- If the limit equals 1, the test is inconclusive.
Root Test
The Root Test, also known as the Radical Test, helps analyze the convergence of a series by examining the limit of the \( k \)-th root of the absolute term itself. Here we find \( \lim_{k \to \infty} (|a_k|)^{1/k} \).
This tells us:
This tells us:
- If this limit is less than 1, the series converges absolutely.
- If greater than 1, the series diverges.
- If it is equal to 1, the test is inconclusive.
Sequences and Series
In mathematics, sequences and series are foundational concepts. A sequence is simply a list of numbers ordered in a specific, logical way. Example: \( a_1, a_2, a_3, \ldots \). Each number in the list corresponds to a natural number index.
A series is the sum of the terms of a sequence. If a sequence \( a_k \) is finite and you add them up, you have a finite series. However, in calculus, we most often work with infinite series, which is the ongoing summation of an infinite sequence \( \sum_{k=1}^{if} a_k \). Understanding when these infinite series converge is crucial. Identifying whether a series converges means determining if the sum approaches a specific number as more of its terms are included.
A series is the sum of the terms of a sequence. If a sequence \( a_k \) is finite and you add them up, you have a finite series. However, in calculus, we most often work with infinite series, which is the ongoing summation of an infinite sequence \( \sum_{k=1}^{if} a_k \). Understanding when these infinite series converge is crucial. Identifying whether a series converges means determining if the sum approaches a specific number as more of its terms are included.
Convergence of Series
The convergence of series tells us if the sum of series results in a finite number. A series \( \sum_{k=1}^{if} a_k \) is said to converge if the sequence of partial sums \( S_n = a_1 + a_2 + \ldots + a_n \) converges to a limit \( S \) as \( n \to \infty \).
There are several methods and tests to determine convergence:
There are several methods and tests to determine convergence:
- Ratio Test is useful for series with factorials or constant powers.
- Root Test is effective for series with exponential growth or roots.
- Other tests include comparison test, integral test, and more.
Other exercises in this chapter
Problem 10
(Ratio Comparison Test) Let \(\left(a_{k}\right)\) and \(\left(b_{k}\right)\) be sequences and suppose \(b_{k}>0\) for all \(k\). Prove the following: (i) If \(
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Let \(a, b \in \mathbb{R}\) be such that \(01\).
View solution Problem 13
Let \(\left(a_{k}\right)\) be a sequence of real numbers. If there is \(p>1\) such that $$ \left|a_{k+1}\right| \leq\left(1-\frac{p}{k}\right)\left|a_{k}\right|
View solution Problem 14
(i) If \(a_{1}:=1\) and \(a_{k+1}:=(k-1) a_{k} /(k+1)\) for \(k \geq 2\), then show that \(\sum_{k=1}^{\infty} a_{k}\) is convergent. (ii) If \(a_{1}:=1\) and \
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