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Chapter 5: Problem 72

Compute the first four partial sums \(S_{1}, \ldots, S_{4}\) for the series having \(n\) th term \(a_{n}\) starting with \(n=1\) as follows. $$ a_{n}=1 / n $$

Short Answer

Expert verified
The first four partial sums are: \( S_1 = 1 \), \( S_2 = \frac{3}{2} \), \( S_3 = \frac{11}{6} \), \( S_4 = \frac{25}{12} \).

Step by step solution

01

Understanding the Series Terms

The series is defined with its general term as \( a_n = \frac{1}{n} \). This means that the first term is \( a_1 = \frac{1}{1} = 1 \), the second term is \( a_2 = \frac{1}{2} \), the third term is \( a_3 = \frac{1}{3} \), and so on.
02

Computing the First Partial Sum, \( S_1 \)

To compute the first partial sum, \( S_1 \), sum the first term in the series. Thus, \( S_1 = a_1 = 1 \).
03

Computing the Second Partial Sum, \( S_2 \)

The second partial sum adds the first two terms in the series. So, \( S_2 = a_1 + a_2 = 1 + \frac{1}{2} = \frac{3}{2} \).
04

Computing the Third Partial Sum, \( S_3 \)

To find the third partial sum, add the first three terms. Thus, \( S_3 = a_1 + a_2 + a_3 = 1 + \frac{1}{2} + \frac{1}{3} = \frac{11}{6} \).
05

Computing the Fourth Partial Sum, \( S_4 \)

The fourth partial sum includes the first four terms. Therefore, \( S_4 = a_1 + a_2 + a_3 + a_4 = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{25}{12} \).

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Key Concepts

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

Series
The concept of a series is a foundational element in mathematics, especially in calculus and analysis. A series is essentially the sum of the terms of a sequence. When you think about a sequence, imagine a list of numbers following a specific rule. If we take these numbers and add them together, we form a series. For example, with the sequence given by the term \( a_n = \frac{1}{n} \), the series becomes the sum of terms like \( 1, \frac{1}{2}, \frac{1}{3} \), and so on. When solving problems with series, mathematicians often look for patterns and convergence.
Harmonic Series
The harmonic series is a special type of series where each term is the reciprocal of an integer. Specifically, it takes the form:\[ a_n = \frac{1}{n} \]. The series begins with 1, \( \frac{1}{2}, \frac{1}{3} \), and so forth. The term "harmonic" originates from historical developments in music and mathematical harmony. This particular series is famous because, although each of its terms seems to get smaller and smaller, the series itself is divergent, meaning it doesn't settle to a definitive limit.
Infinite Series
An infinite series is essentially an extension of the finite series concept. It keeps going indefinitely, adding up terms forever, without a firm endpoint. An infinite series can be written symbolically with the summation notation \( \sum_{n=1}^{\infty} a_n \), where \( n \) starts at 1 and theoretically goes on to infinity. In practical scenarios, we analyze infinite series to see if they converge or diverge. The crucial point is whether the total sum approaches a finite value (converges) or not (diverges). Various tests help determine this, like the comparison test and the ratio test.
Convergence
Convergence in the context of series and sequences refers to whether a series approaches a given limit as more terms are added. For a series like the harmonic series, even though it has an infinite number of terms, we might inquire if this sum gets closer and closer to a number. If it does, we say the series converges; if not, it diverges. Interestingly, the harmonic series is a classic example of a divergent series, despite its terms growing smaller. Understanding convergence is vital as it helps predict the behavior of series and determine if calculations can lead to logical, finite results.

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Most popular questions from this chapter

[T] Estimates of \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\) can be accelerated by its \(\quad\) partial \(\quad\) sums writing \(\quad\) its \(\quad\) partial \(\sum_{n=1}^{N} \frac{1}{n^{2}}=\sum_{n=1}^{N} \frac{1}{n(n+1)}+\sum_{n=1}^{N} \frac{1}{n^{2}(n+1)}\) and recalling that \(\sum_{n=1}^{N} \frac{1}{n(n+1)}=1-\frac{1}{N+1}\) converges to one as \(N \rightarrow \infty\). Compare the estimate of \(\pi^{2} / 6\) using the sums \(\sum_{n=1}^{1000} \frac{1}{n^{2}}\) with the estimate using \(1+\sum_{n=1}^{1000} \frac{1}{n^{2}(n+1)} .\)

In the following exercises, use an appropriate test to determine whether the series converges. $$a_{k}=1 /\left(\begin{array}{l}2 k \\ k\end{array}\right)$$

Let \(a_{n}=2^{-[n / 2]}\) where \([x]\) is the greatest integer less than or equal to \(x\). Determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges and justify your answer. The following advanced exercises use a generalized ratio test to determine convergence of some series that arise in particular applications when tests in this chapter, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if \(\lim _{n \rightarrow \infty} \frac{a_{2 n}}{a_{n}}<1 / 2\), then \(\sum a_{n}\) converges, while if \(\lim _{n \rightarrow \infty} \frac{a_{2 n+1}}{a_{n}}>1 / 2\), then \(\sum a_{n}\) diverges.

Use the root test to determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is as follows. $$a_{k}=\frac{k^{e}}{e^{k}}$$

Use the root test to determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is as follows. $$a_{n}=\frac{(\ln n)^{2 n}}{n^{n}}$$
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