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Converge absolutely, which converge, and which diverge? Give reasons for your answers. $$\sum_{n=1}^{\infty}(-1)^{n+1}(0.1)^{n}$$

Short Answer

Expert verified
The series converges absolutely because its absolute series is a convergent geometric series.

Step by step solution

01

Identify the Series

The given series is \( \sum_{n=1}^{\infty} (-1)^{n+1} (0.1)^n \). This is an alternating series because of the \((-1)^{n+1}\) term.
02

Check Absolute Convergence

To check for absolute convergence, we consider the absolute series \( \sum_{n=1}^{\infty} |(-1)^{n+1} (0.1)^n| = \sum_{n=1}^{\infty} (0.1)^n \). This is a geometric series with a common ratio \( r = 0.1 \), which satisfies \( |r| < 1 \). A geometric series converges if its common ratio's absolute value is less than 1.
03

Apply the Geometric Series Test

Since \( |r| = 0.1 < 1 \), the series \( \sum_{n=1}^{\infty} (0.1)^n \) converges. Therefore, the original series converges absolutely.
04

Conclusion

An absolutely convergent series converges. Therefore, the series \( \sum_{n=1}^{\infty} (-1)^{n+1} (0.1)^n \) converges absolutely.

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Key Concepts

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

Absolute Convergence
Absolute convergence is when the series of absolute values of terms converges. In simpler words, if you strip away the signs (like the negative in an alternating series) and the series still converges, then your original series converges absolutely.

Think of it like this: Absolute convergence is a stronger form of convergence. When you have a series, say \( \sum_{n=1}^{\infty} a_n \), you check if \( \sum_{n=1}^{\infty} |a_n| \) converges. If it does, then you have absolute convergence.
  • Absolutely convergent series are always convergent.
  • The reverse isn't always true - convergent series are not always absolutely convergent.
In the problem given, the series \( \sum_{n=1}^{\infty} |(-1)^{n+1} (0.1)^n| \) turns into a geometric series with a positive \((0.1)^n\), which we test further for convergence. If it's convergent, that means the original series converges absolutely.
Geometric Series
A geometric series is one where each term is a constant multiple, known as the common ratio, of the previous term. It is like stepping stones that always take you an equal distance from the last one.

To tell if a geometric series converges, look at its common ratio, \(r\). The rule is straightforward:
  • If \(|r| < 1\), the series converges.
  • If \(|r| \geq 1\), the series diverges.
The series in our problem, \(\sum_{n=1}^{\infty} (0.1)^n\), has a common ratio \(0.1\). Given \(|0.1| = 0.1 < 1\), the series converges according to the geometric series test. This means our geometric series "passes" and helps confirm absolute convergence of the original series.
Series Convergence Analysis
Analyzing series convergence requires piecing together several tests, not just relying on intuition. It is like using different maps to understand the terrain better.

To analyze series convergence, you start by selecting the most appropriate test.
  • For alternating series, the Alternating Series Test might be used.
  • For positive term series, the Ratio Test or Geometric Series Test often applies.
Each test plays a role like a compass for different conditions. In our exercise, first turning the series into its absolute form allowed us to see it as a geometric series, which we could then test for convergence directly.

It's important always to verify which test suits the specific series. By understanding the convergence through the geometric series approach, and confirming absolute convergence, we conclude the regularly changing signs in terms (due to the alternating factor) don’t affect this conclusion drastically.

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Most popular questions from this chapter

Sequences generated by Newton's method \(\quad\) Newton's method, applied to a differentiable function \(f(x),\) begins with a starting value \(x_{0}\) and constructs from it a sequence of numbers \(\left\\{x_{n}\right\\}\) that under favorable circumstances converges to a zero of \(f .\) The recursion formula for the sequence is $$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$ a. Show that the recursion formula for \(f(x)=x^{2}-a, a>0\) can be written as \(x_{n+1}=\left(x_{n}+a / x_{n}\right) / 2\). b. Starting with \(x_{0}=1\) and \(a=3,\) calculate successive terms of the sequence until the display begins to repeat. What number is being approximated? Explain.

The estimate \(\sqrt{1+x}=1+(x / 2)\) is used when \(x\) is small. Estimate the error when \(|x|<0.01\).

a. Use Taylor's formula with \(n=2\) to find the quadratic approximation of \(f(x)=(1+x)^{k}\) at \(x=0\) ( \(k\) a constant). b. If \(k=3,\) for approximately what values of \(x\) in the interval [0,1] will the error in the quadratic approximation be less than \(1 / 100 ?\)

A patient takes a 300 mg tablet for the control of high blood pressure every morning at the same time. The concentration of the drug in the patient's system decays exponentially at a constant hourly rate of \(k=0.12\) a. How many milligrams of the drug are in the patient's system just before the second tablet is taken? Just before the third tablet is taken? b. In the long run, after taking the medication for at least six months, what quantity of drug is in the patient's body just before taking the next regularly scheduled morning tablet?

a. Suppose that \(f(x)\) is differentiable for all \(x\) in [0,1] and that \(f(0)=0 .\) Define sequence \(\left\\{a_{n}\right\\}\) by the rule \(a_{n}=n f(1 / n)\) Show that \(\lim _{n \rightarrow \infty} a_{n}=f^{\prime}(0) .\) Use the result in part (a) to find the limits of the following sequences \(\left\\{a_{n}\right\\}\). b. \(a_{n}=n \tan ^{-1} \frac{1}{n}\) c. \(a_{n}=n\left(e^{1 / n}-1\right)\) d. \(a_{n}=n \ln \left(1+\frac{2}{n}\right)\)

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