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Chapter 9: Problem 3

Show that the series \(\sum_{k \geq 1} 2 /(k+1)(2 k+1)\) is convergent and its sum is less than or equal to 1 . (Hint: Compare the given series with the series \(\left.\sum_{k \geq 1} 1 / k(k+1) .\right)\)

Short Answer

Expert verified
The series \(\sum_{k \geq 1} \frac{2}{(k+1)(2k+1)}\) converges and its sum is less than or equal to 1. We used the comparison test by comparing it to the series \(\sum_{k \geq 1} \frac{1}{k(k+1)}\), for which we computed partial sums \(S_n = 1 - \frac{1}{n+1}\). Since the partial sums of the given series \(T_n\) are bounded by the partial sums of the convergent sequence \(S_n\), the given series also converges and its sum is less than or equal to \(\lim_{n\to\infty} S_n = 1\).

Step by step solution

01

Compare the series using the comparison test

First, note that for all \(k \geq 1\), the following inequality holds: \(\frac{1}{k(k+1)} \leq \frac{2}{(k+1)(2k+1)}\) This is true because the denominator of the former is smaller than the latter, so the whole fraction is greater. Now, let's compute partial sums of the series \(\sum_{k \geq 1} \frac{1}{k(k+1)}\), that we know converges, to use it in the comparison test.
02

Compute partial sums of the series \(\sum_{k \geq 1} \frac{1}{k(k+1)}\)

We will now compute the partial sum of the series \(\sum_{k \geq 1} \frac{1}{k(k+1)}\). We can rewrite the general term as follows: \(\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}\) Now, we will compute the partial sum for the first n terms: \(S_n = \sum_{k=1}^{n}\left(\frac{1}{k} - \frac{1}{k+1}\right) = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)\) Notice that most terms cancel to leave only: \(S_n = 1 - \frac{1}{n+1}\)
03

Use the comparison test to show the given series converges

Now, we can use the partial sum of the series \(\sum_{k \geq 1} \frac{1}{k(k+1)}\) to show that the given series converges. Let \(T_n\) be the partial sum of the first n terms of the given series, we have: \(T_n \leq S_n\) Because of the inequality \(\frac{1}{k(k+1)} \leq \frac{2}{(k+1)(2k+1)}\) we showed in step 1 holds for all \(k \geq 1\). As n goes to infinity, we have: \(\lim_{n\to\infty} S_n = \lim_{n\to\infty} \left(1 - \frac{1}{n+1}\right) = 1\) Since \(T_n\) is bounded by a converging sequence, then by the comparison test, the given series also converges.
04

Show that the sum is less than or equal to 1

To show that the sum of the given series is less than or equal to 1, we can use the partial sums \(T_n\). Since \(T_n \leq S_n\), then: \(\lim_{n\to\infty} T_n \leq \lim_{n\to\infty} S_n\) As we showed earlier, \(\lim_{n\to\infty} S_n = 1\), thus: \(\lim_{n\to\infty} T_n \leq 1\) Which means that the sum of the given series is less than or equal to 1. In conclusion, the series \(\sum_{k \geq 1} \frac{2}{(k+1)(2k+1)}\) is convergent and its sum is less than or equal to 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Comparison Test
When it comes to understanding the convergence of a given series, the comparison test is an invaluable tool. This method entails comparing the series in question to another series whose convergence properties are well understood. The key principle behind the comparison test is that if the terms of one series are always smaller (or larger) than the corresponding terms of a series that converges (or diverges), then the original series will also converge (or diverge).

For example, in the exercise provided, the series \(\sum_{k \geq 1} 2 /(k+1)(2 k+1)\) is compared to \(\sum_{k \geq 1} 1 / k(k+1) \). By showing that one term is less than or equal to the other and that the series with the lesser terms converges, we can conclude through the comparison test that the series in question also converges. This technique simplifies the work of proving convergence, allowing students to build upon known series behaviors to tackle new problems.
Partial Sums
Partial sums play a crucial role in determining the convergence of a series. These are simply the sums of the first 'n' terms of the series, denoted as \(S_n\). To get the nth partial sum, we sequentially add terms from the first up to the nth term of the series. The behavior of the partial sums as n approaches infinity鈥攖o what value, if any, they settle鈥攔eveals whether the series diverges or converges.

In our exercise, the partial sum \(S_n\) is calculated by summing the terms \(\frac{1}{k} - \frac{1}{k+1}\), resulting in a series of cancellations that leave behind the expression \(S_n = 1 - \frac{1}{n+1}\). By examining these partial sums, we can predict the behavior of the entire series, which is pivotal to proving convergence using the comparison test.
Limit of a Sequence
The limit of a sequence is a fundamental concept in calculus that describes the value the terms of a sequence approach as the index increases indefinitely. If the sequence has a finite limit, we say that the sequence converges, otherwise it diverges. The concept of the limit is tightly intertwined with the notion of series convergence because the limit of the partial sums' sequence determines the series' fate.

In simple terms, if the limit of the partial sums of a series \(\lim_{n\to\infty} S_n\) exists and is finite, then the series converges to that limit. This is perfectly illustrated in the exercise solution where \(\lim_{n\to\infty} S_n = 1\), affirming the convergence of the series used for comparison.
Series Convergence
Series convergence is an intuitively gripping yet technically intricate part of mathematics. A series is said to converge if the sequence of its partial sums tends to a specific limit. Therefore, to determine the convergence of a series, one must investigate the behavior of its partial sums as the number of terms grows indefinitely.

Using tools like the comparison test, limits of sequences, and the computation of partial sums, as demonstrated in the provided exercise, we get a thorough understanding of the series' behavior. The concluding analysis in the exercise shows that the series \(\sum_{k \geq 1} \frac{2}{(k+1)(2k+1)}\) not only converges but does so to a sum that is less than or equal to 1. These conclusions are pivotal in many areas of math, physics, and engineering where understanding the sum of an infinite series can describe phenomena or predict outcomes practically.

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Most popular questions from this chapter

(Grouping of Terms) Let \(m_{0}:=0\) and let \(m_{1}1} a_{k}\) is convergent, then show that the series $\sum_{k \geq 1} b_{k}$ is convergent and has the same sum. Give an example to show that \(\sum_{k>1} b_{k}\) may be convergent although \(\sum_{k>1} a_{k}\) is divergent.

Let \(f, g:[2, \infty) \rightarrow \mathbb{R}\) be defined by $$ f(t):=\left\\{\begin{array}{ll} 1 & \text { if } k \leq t

Suppose one of the series \(\sum_{k \geq 0} a_{k}\) and \(\sum_{k \geq 0} b_{k}\) is absolutely convergent and the other is convergent. Let \(A\) and \(B\) denote their respective sums. For each \(k=0,1, \ldots\), let \(c_{k}:=\sum_{j=0}^{k} a_{j} b_{k-j} .\) Show that the series \(\sum_{k \geq 0} c_{k}\) is convergent and its sum is equal to \(A B\). Give an example to show that the result may not hold if both \(\sum_{k \geq 0} a_{k}\) and \(\sum_{k \geq 0} b_{k}\) are conditionally convergent. [Note: If \(\sum_{k \geq 0} a_{k}\) and \(\sum_{k \geq 0} b_{k}\) are convergent, and if the series \(\sum_{k \geq 0} c_{k}\) is convergent, then its sum must be \(A B\). This can be proved using a result of Abel given in Exercise \(10.44\) (i).]

Let \(f:[1, \infty) \rightarrow \mathbb{R}\) be such that \(f\) is integrable on \([1, x]\) for every \(x \geq 1\). Prove the following using Proposition 9.46: (i) If there exist \(p>1\) and \(\ell \in \mathbb{R}\) such that \(t^{p} f(t) \rightarrow \ell\) as \(t \rightarrow \infty\), then \(\int_{t \geq 1} f(t) d t\) is absolutely convergent. (ii) Suppose \(f(t)>0\) for all \(t \in[1, \infty)\). If there exist \(p \leq 1\) and \(\ell \neq 0\) such that \(t^{p} f(t) \rightarrow \ell\) as \(t \rightarrow \infty\), then \(\int_{t \geq 1} f(t) d t\) is divergent.

Show that \(\int_{t \geq 1}\left(\cos t / t^{p}\right) d t\) and \(\int_{t \geq 1}\left(\sin t / t^{p}\right) d t\) are absolutely convergent if \(p>1\) and that they are conditionally convergent if \(0
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