Problem 2 It is useful to write series bot... [FREE SOLUTION]

Chapter 1: Problem 2

It is useful to write series both in the form $a_{1}+a_{2}+a_{3}+\cdots$ and in the form $\sum_{n=1}^{\infty} a_{n} .$ Write out several terms of the following series (that is, write them in the first form). $$ \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} $$

Short Answer

Expert verified

-1 + $\frac{1}{2}$ - $\frac{1}{3}$ + $\frac{1}{4}$ + \cdots\

Step by step solution

Understand the Series Notation

The given series is represented in summation notation as $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}$\. This notation means that you add each term of the form $\frac{(-1)^{n}}{n}$\, starting from $n = 1$ and continuing indefinitely.

Write the First Term

To find the first term of the series, substitute $n = 1$ into the general term $\frac{(-1)^{n}}{n}$\: \[ \frac{(-1)^{1}}{1} = -1 \]

Write the Second Term

Next, substitute $n = 2$ into the general term: \[ \frac{(-1)^{2}}{2} = \frac{1}{2} \]

Write the Third Term

Substitute $n = 3$ into the general term: \[ \frac{(-1)^{3}}{3} = -\frac{1}{3} \]

Write the Fourth Term

Substitute $n = 4$ into the general term: \[ \frac{(-1)^{4}}{4} = \frac{1}{4} \]

Combine the Terms

Combine the first four terms to write out the series in its expanded form: \[ -1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \cdots \]

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Key Concepts

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

summation notation

Summation notation uses the symbol $\sum$ to represent the sum of a sequence of terms. It's a concise way to express long sums. The notation includes an index of summation, typically $n$, which starts at a lower limit and goes up to an upper limit. For example, $\sum_{n=1}^{\infty} a_{n}$ means you are summing the terms $a_n$ starting from $n=1$ and continuing indefinitely. This can efficiently describe infinite series and helps in understanding the behavior of sums.

general term

In a series, the general term is an expression that represents any term in the sequence, based on its position $n$. For the series given: $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}$, the general term is $\frac{(-1)^{n}}{n}$. By plugging different values of $n$ into this formula, you can find specific terms in the series.
For example:

When $n = 1$: $\frac{(-1)^{1}}{1} = -1$
When $n = 2$: $\frac{(-1)^{2}}{2} = \frac{1}{2}$
When $n = 3$: $\frac{(-1)^{3}}{3} = -\frac{1}{3}$
When $n = 4$: $\frac{(-1)^{4}}{4} = \frac{1}{4}$

Combining these general terms helps in writing the series in its expanded form.

alternating series

An alternating series is a series where the signs of the terms alternate between positive and negative. This is caused by having a factor like $(-1)^n$ or $(-1)^{n+1}$ in the general term.
The series $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}$ is an alternating series because the $(-1)^n$ factor changes the sign of each term.

For $n=1$, the term is $-1$
For $n=2$, the term is $\frac{1}{2}$
For $n=3$, the term is $-\frac{1}{3}$
and so on.

This alternating nature often affects the convergence and behavior of the series, making it an interesting topic in studies of convergence tests.

infinite series

An infinite series is a sum of infinitely many terms. Instead of stopping at a terminal number, it continues forever. For example, $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}$ is an infinite series as it goes on without end.
Important aspects of infinite series include:

**Convergence**: Does the series approach a specific value as more terms are added?
**Divergence**: Does the series fail to approach a fixed value?

Infinite series can represent many mathematical functions and phenomena. They require specific techniques to analyze their behavior, such as the Ratio Test, Root Test, or Integral Test, which help determine whether a series converges or diverges.

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91影视

It is useful to write series both in the form \(a_{1}+a_{2}+a_{3}+\cdots\) and in the form \(\sum_{n=1}^{\infty} a_{n} .\) Write out several terms of the following series (that is, write them in the first form). $$ \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} $$