Problem 1
Question
Investigate the convergence of the following series: (a) \(\frac{1}{3}+\frac{2}{6}+\frac{3}{11}+\frac{4}{18}+\frac{5}{27}+\cdots\) (b) \(\frac{1}{2}-\frac{2}{20}+\frac{3}{38}-\frac{4}{56}+\frac{5}{74}-\cdots\) (c) \(\frac{1}{3}+\frac{1 \cdot 2}{3 \cdot 5}+\frac{1 \cdot 2 \cdot 3}{3 \cdot 5 \cdot 7}+\cdots\) (d) \(\frac{1}{4}+\frac{1 \cdot 9}{4 \cdot 16}+\frac{1 \cdot 9 \cdot 25}{4 \cdot 16 \cdot 36}+\frac{1 \cdot 9 \cdot 25 \cdot 49}{4 \cdot 16 \cdot 36 \cdot 64}+\cdots\)
Step-by-Step Solution
Verified Answer
(a) The series converges. (b) The series converges.
1Step 1: Define the general term for (a)
The general term for the series in (a) can be observed as \(a_n=(-1)^n*\frac{n}{2n-1}\) for \(n \geq 1\).
2Step 2: Apply the Alternating series test for (a)
The alternating series test states that if two conditions are met: 1) \(a_{n+1}\) is smaller than or equal to \(a_n\) for all \(n \geq 1\), and also \(a_n\) approaches zero as \(n \to \infty\), the series \(-1^n a_n\) converges. For our series, we can see that the sequence \(a_n\) is positive and decreasing, and \(a_n\) approaches 0 as \(n \to \infty\). Therefore, the series converges.
3Step 3: Define the general term for (b)
The general term has alternating signs, and the elements in the denominator increment by two every two terms. This could be expressed as \(b_n = (-1)^{(n-1)/2}\frac{1}{\sqrt{n+(n-1)/2}}\) for \(n \geq 1\). However, this formula can be difficult to handle. Instead, notice that every second term is negative and the denominator of these terms is a perfect square. Therefore, it's useful to split the sequence into two subsequences.
4Step 4: Split (b) into two subsequences
Split (b) into two subsequences: one containing positive terms, say \(b_{n1}= 1/\sqrt{2n+1}\) for \(n \geq 1\) and another containing negative terms, say \(b_{n2}=-1/\sqrt{2n}\) for \(n \geq 1\).
5Step 5: Apply Limit comparison test for (b)
For both subsequences of (b), we can apply the limit comparison test with \(1/n^2\) for positive series and with \(1/n\) for negative series. Given \(b_{n1}\) and \(b_{n2}\), we can see that both subsequences converge following the p-series test. Hence, the original series is the sum of two convergent series and thus converges.
Key Concepts
Alternating Series TestLimit Comparison TestP-Series Test
Alternating Series Test
Understanding the Alternating Series Test is crucial when dealing with series whose terms alternate between positive and negative. This test is applied to determine if an infinite series of the form \(\sum (-1)^{n} a_{n}\) is convergent.
For the test to be valid, two main conditions must be met. Firstly, the absolute values of the series terms \(a_n\) must be decreasing; in other words, \(a_{n+1} \leq a_n\) for all \(n\). Secondly, the terms must approach zero as \(n\) approaches infinity \(\lim_{n \to \infty} a_n = 0\).
When both conditions are satisfied, the series is guaranteed to converge. Let's examine series (a) from our exercise. By defining the general term as \(a_n=(-1)^n*\frac{n}{2n-1}\) for \(n \geq 1\), it's evident that \(a_n\) is positive, decreasing, and approaches zero, satisfying the criteria of the Alternating Series Test. Hence, series (a) converges.
For the test to be valid, two main conditions must be met. Firstly, the absolute values of the series terms \(a_n\) must be decreasing; in other words, \(a_{n+1} \leq a_n\) for all \(n\). Secondly, the terms must approach zero as \(n\) approaches infinity \(\lim_{n \to \infty} a_n = 0\).
When both conditions are satisfied, the series is guaranteed to converge. Let's examine series (a) from our exercise. By defining the general term as \(a_n=(-1)^n*\frac{n}{2n-1}\) for \(n \geq 1\), it's evident that \(a_n\) is positive, decreasing, and approaches zero, satisfying the criteria of the Alternating Series Test. Hence, series (a) converges.
Limit Comparison Test
Next, we delve into the Limit Comparison Test, which serves as a powerful tool for determining the convergence of series by comparing them with a known convergent or divergent series.
To use this test, take the series of interest, \(\sum b_n\), and a second, comparison series, \(\sum a_n\), that you already know converges or diverges. The test involves evaluating the limit \(\lim_{n \to \infty}\frac{b_n}{a_n}\). If this limit is a positive finite number, both series will converge or diverge together.
In the case of series (b), by splitting the series into two subsequences \(b_{n1}= \frac{1}{\sqrt{2n+1}}\) and \(b_{n2}=-\frac{1}{\sqrt{2n}}\), and comparing each with the convergent p-series \(\frac{1}{n^2}\) and \(\frac{1}{n}\) respectively, we find that both limits are positive finite numbers. This indicates that both subsequences, and consequently series (b), convergent.
To use this test, take the series of interest, \(\sum b_n\), and a second, comparison series, \(\sum a_n\), that you already know converges or diverges. The test involves evaluating the limit \(\lim_{n \to \infty}\frac{b_n}{a_n}\). If this limit is a positive finite number, both series will converge or diverge together.
In the case of series (b), by splitting the series into two subsequences \(b_{n1}= \frac{1}{\sqrt{2n+1}}\) and \(b_{n2}=-\frac{1}{\sqrt{2n}}\), and comparing each with the convergent p-series \(\frac{1}{n^2}\) and \(\frac{1}{n}\) respectively, we find that both limits are positive finite numbers. This indicates that both subsequences, and consequently series (b), convergent.
P-Series Test
Finally, let's examine the P-Series Test, which is a simple yet effective criterion for testing the convergence of series of the form \(\sum \frac{1}{n^p}\).
The test states that this type of series converges if and only if the exponent \(p>1\). Conversely, if \(p \leq 1\), the series diverges. This is because when \(p>1\), the terms of the series decrease rapidly enough to ensure the sum does not grow infinitely large.
In series (b) from our exercise, after splitting it, we used the P-Series Test by comparing with \(\frac{1}{n^2}\) and \(\frac{1}{n}\), which are known p-series. Since the exponents in the comparators are 2 and 1, corresponding to convergent and divergent series, and the limits we found in the Limit Comparison Test were finite and positive, we conclude that the positive and negative subsequences converge. The convergence of these subsequences infers that series (b) itself is convergent.
The test states that this type of series converges if and only if the exponent \(p>1\). Conversely, if \(p \leq 1\), the series diverges. This is because when \(p>1\), the terms of the series decrease rapidly enough to ensure the sum does not grow infinitely large.
In series (b) from our exercise, after splitting it, we used the P-Series Test by comparing with \(\frac{1}{n^2}\) and \(\frac{1}{n}\), which are known p-series. Since the exponents in the comparators are 2 and 1, corresponding to convergent and divergent series, and the limits we found in the Limit Comparison Test were finite and positive, we conclude that the positive and negative subsequences converge. The convergence of these subsequences infers that series (b) itself is convergent.
Other exercises in this chapter
Problem 1
Since \(\sum_{2}^{\infty} 1 /(n \log n)\) diverges, \(\lim _{n \rightarrow \infty} \sum_{2}^{n} 1 /(k \log k)=\infty .\) How many terms must be taken before the
View solution Problem 2
Form the Cauchy product of the following series: $$ \begin{aligned} &1+2+4+8+16+32+\cdots \\ &1-1+1-1+1-1+1-1+\cdots \end{aligned} $$ and find a formula for the
View solution Problem 2
Show that if \(\sum a_{n}\) converges, then \(\sum_{N}^{\infty} a_{n} \rightarrow 0\) as \(N \rightarrow \infty\)
View solution Problem 3
Find a formula for the coefficients of the Cauchy product of the series \(\sum_{0}^{\infty} A^{n}\) and \(\sum_{0}^{\infty} B^{n} .\)
View solution