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Chapter 9: Problem 2

Compute the fourth partial sum of each series. $$ \sum_{n=1}^{\infty} n $$

Short Answer

Expert verified
The fourth partial sum is 10.

Step by step solution

01

Understand the Partial Sum

The partial sum of a series is the sum of a finite number of its terms. Here, you are asked to compute the fourth partial sum of the series which is written as \( \sum_{n=1}^{\infty} n \). This series lists all natural numbers: 1, 2, 3, 4, 5, and so on. For the fourth partial sum, you will sum the first four terms of the series.
02

Identify the First Four Terms

The first four terms of the series \( \sum_{n=1}^{\infty} n \) are 1, 2, 3, and 4. These are the terms you will add together to find the fourth partial sum.
03

Add the First Four Terms

Now, add the first four terms: \( 1 + 2 + 3 + 4 \). Calculating this results in: \( 1 + 2 = 3 \), \( 3 + 3 = 6 \), \( 6 + 4 = 10 \). The sum is 10.

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Key Concepts

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

Infinite Series
An infinite series is a sum of an unending sequence of terms. In mathematics, you often encounter this kind of series when dealing with sums that do not terminate. Instead of having a finite number of terms, they go on forever. Not all infinite series converge to a specific number. This means they don't always add up to a value you can nail down. However, some infinite series do converge and can be quite useful.
  • Common examples are geometric and harmonic series.
  • The series presented in our example is an arithmetic series of all natural numbers.
When you're asked to find a partial sum, you're looking for the sum of the first few terms rather than the "whole" series, which is really endless. It's a way of diving into an infinite series and capturing just a piece of it.
Natural Numbers
Natural numbers are the numbers you naturally start counting with: 1, 2, 3, 4, and so forth. They are the positive integers. When you see a series like the one in our problem, the terms are made up of these natural numbers.
  • They are also called whole numbers, although some definitions may include 0 in whole numbers, natural numbers typically begin from 1.
  • They form the basis for counting and ordering.
Natural numbers are simple yet fundamental, appearing everywhere in mathematics. In our situation, summing them helps us compute partial sums of a series. Understanding this concept lays a strong foundation for grasping more complex mathematical ideas.
Arithmetic Series
An arithmetic series emerges from adding the terms of an arithmetic sequence. The sequence involves increasing each term by a constant amount called the common difference. For example, in the series given in the exercise, the common difference is 1.
To compute the partial sums of an arithmetic series:
  • Identify the common difference between terms.
  • Add the sequence of numbers starting from the first term.
In the exercise, you'll find that the fourth partial sum was reached by adding the terms 1, 2, 3, and 4, totaling 10. Mastering arithmetic series enables you to manage sums systematically, by recognizing patterns and building from fundamental principles. It's a technique applicable to both simple and more intricate problems alike.

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Most popular questions from this chapter

Find the Taylor series of the given function about \(a\). Use the series already obtained in the text or in previous exercises. $$ g(x)=\sinh x ; a=0 $$

Let \(a \neq 0\), and assume that \(\lim _{n \rightarrow \infty} a_{n}=a\) and \(a_{n} \neq 0\) for all \(n\). Show that \(\sum_{n=1}^{\infty}\left|a_{n+1}-a_{n}\right|\) converges if and only if \(\sum_{n=1}^{\infty}\left|\frac{1}{a_{n+1}}-\frac{1}{a_{n}}\right|\) converges.

Let \(f(x)\) be the sum of the series. Find \(f^{\prime}(x)\) and \(\int_{0}^{x} f(t) d t\). $$ \sum_{n=1}^{\infty}(n+1) x^{n} $$

Find the Taylor series of \(f\) about \(a\), and write out the first four terms of the series. $$ f(x)=\left(1-x^{2}\right)^{5 / 2} ; a=0 $$

Determine which series diverge, which converge conditionally, and which converge absolutely. $$ \sum_{n=3}^{\infty}(-1)^{n+1} \frac{1}{n(n-2)} $$
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