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Chapter 9: Problem 12

Let \(a>0\). Show that the series \(\sum\left(1+a^{n}\right)^{-1}\) is divergent if \(01\)

Short Answer

Expert verified
The series \(\sum\left(1+a^{n}\right)^{-1}\) is divergent if \(01\).

Step by step solution

01

Case 1: For \(0

For \(0<a \leq 1\), let's compare the series with \(\sum\left(1+n\right)^{-1}\), which is a harmonic series. This harmonic series is known to be divergent. We have \(1/(1+a^n) \geq 1/(1+n)\) because \(a^n \leq n\). If \(0<a \leq 1\), then \(\sum\left(1+a^{n}\right)^{-1}\) is a series with terms greater than or equal to the terms of the harmonic series. Therefore, by the comparison test, the series \(\sum\left(1+a^{n}\right)^{-1}\) is also divergent when \(0<a \leq 1\).
02

Case 2: For \(a>1\)

For \(a>1\), let's apply the root test to the series. The root test focuses on the limit of the nth root of the absolute value of each term in the series. Here, we compute the limit as \(n\) goes to infinity of \((1+a^n)^{-1/n}\), which equals \(a^{-1}\) because the \(n\)th root of \(1\) is \(1\) and the \(n\)th root of \(a^n\) is \(a\). As \(a>1\), \(a^{-1}<1\). The root test states that if this limit is less than \(1\), the series converges. Therefore, the series \(\sum\left(1+a^{n}\right)^{-1}\) is convergent when \(a>1\).

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Key Concepts

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

Comparison Test
The Comparison Test is a technique used to determine the convergence or divergence of an infinite series by comparing it to another series with known behavior. In simple terms, if you can show that the series you are investigating is larger than a divergent series, then it too must diverge. Conversely, if it is smaller than a convergent series, it must converge.

To apply this test:
  • Choose a comparison series with terms that are comparable to your series. This means they must be larger or smaller but in some way similar.
  • Make sure the comparison series has a known convergence behavior.
  • If each term of your series is larger than the corresponding term of a divergent series, your series also diverges.
  • If each term is smaller than that of a convergent series, your series converges.
The comparison test is especially useful when dealing with series that resemble basic ideas like the harmonic series.
Harmonic Series
The Harmonic Series is one of the most well-known divergent series in mathematics. It is expressed as \[\sum \frac{1}{n},\] and its terms go on indefinitely: \(\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \ldots\)

Despite the terms approaching zero, the harmonic series diverges. This is a key point that often surprises new learners. The intuition here is that the terms decrease very slowly, so when summed indefinitely, they do not settle to a fixed number.

When using the harmonic series in conjunction with the comparison test, you often look for series with similar term structures that you can confidently say are larger in magnitude.
Root Test
The Root Test, also known as Cauchy's Root Test, is a powerful method used to determine the convergence of a series using the nth root of its terms. It's particularly useful when the terms of the series include exponential expressions.

To apply the Root Test:
  • Calculate the nth root of the absolute value of the terms: \[\lim_{n \to \infty} \sqrt[n]{|a_n|}.\]
  • Determine the limit of this expression as \(n\) approaches infinity.
  • If the limit is less than 1, the series converges.
  • If the limit is greater than 1, the series diverges.
  • If the limit equals 1, the test is inconclusive.
The Root Test is precise for handling series that appear to have a geometric progression hidden within their structure, such as terms that include powers or exponential functions. It grants insight into the series behavior by effectively steering the series towards known convergence territory.

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Most popular questions from this chapter

Let \(\left\\{n_{1}, n_{2}, \cdots\right\\}\) denote the collection of natural numbers that do not use the digit 6 in their decimal expansion. Show that \(\sum 1 / n_{k}\) converges to a number less than 80 . If \(\left\\{m_{1}, m_{2}, \cdots\right\\}\) is the collection of numbers that end in 6 , then \(\sum 1 / m_{k}\) diverges. If \(\left\\{p_{1}, p_{2}, \cdots\right.\), ) is the collection of numbers that do not end in 6 , then \(\sum 1 / p_{k}\) diverges.

If \(\sum a_{n}\) is absolutely convergent, is it true that every rearrangement of \(\sum a_{n}\) is also absolutely convergent?

Prove by Induction that the function given by \(f(x):=e^{-1 / x^{2}}\) for \(x \neq 0, f(0):=0\), has derivatives of all orders at every point and that all of these derivatives vanish at \(x=0\). Hence this function is not given by its Taylor expansion about \(x=0\).

Find a scries expansion for \(\int_{0}^{x} e^{-t^{2}} d t\) for \(x \in \mathbb{R}\).

Let \(a_{n} \in \mathbb{R}\) for \(n \in \mathbb{N}\) and let \(p
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