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Discuss the convergence or divergence of $$ \begin{aligned} \frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}+\\\ \frac{1}{\sqrt{4}-1}-\frac{1}{\sqrt{4}+1}+\cdots \end{aligned} $$

Short Answer

Expert verified
The series diverges.

Step by step solution

01

Identify the General Term

First, let's identify what each term in the series looks like. The series is given by: \[\sum_{n=2}^{\infty} \left( \frac{1}{\sqrt{n}-1} - \frac{1}{\sqrt{n}+1} \right)\]This is the series we must analyze for convergence or divergence.
02

Simplify the General Term

Simplify the expression for the general term:\[\frac{1}{\sqrt{n}-1} - \frac{1}{\sqrt{n}+1} = \frac{(\sqrt{n}+1) - (\sqrt{n}-1)}{(\sqrt{n}-1)(\sqrt{n}+1)}\]\[= \frac{2}{n-1}\]So, the general term simplifies to \( \frac{2}{n-1} \).
03

Examine the Simplified Series

The series then becomes:\[2 \sum_{n=2}^{\infty} \frac{1}{n-1} = 2 \sum_{m=1}^{\infty} \frac{1}{m}\]Where we made the substitution \( m = n-1 \). This is a version of the harmonic series.
04

Determine Convergence or Divergence

We know that the harmonic series, \( \sum_{m=1}^{\infty} \frac{1}{m} \), diverges. Since our series is essentially twice the harmonic series:\[2 \sum_{m=1}^{\infty} \frac{1}{m},\] it also diverges.

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Key Concepts

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

Harmonic Series
The harmonic series is a simple, yet fundamental, series in mathematics. It has the form \(\sum_{n=1}^{\infty} \frac{1}{n} .\) Each term in the harmonic series is the reciprocal of a positive integer. Although these terms get smaller as \( n \) increases, they do not shrink fast enough to make the series converge.
In terms of behavior, the harmonic series is famous for its divergence. Unlike a convergent series, where the sum approaches a specific number, the harmonic series grows without bound. As you include more terms, the sum of the harmonic series increases indefinitely. Understanding the harmonic series is crucial when analyzing other series that resemble it, such as the altered series in this exercise.
General Term Simplification
General term simplification is a technique used to make complex series more manageable and easier to analyze. In this problem, each term in the series is expressed as:\(\frac{1}{\sqrt{n}-1} - \frac{1}{\sqrt{n}+1}.\) To simplify, we combine these fractions over a common denominator:
  • Recognize that both terms share a similar structure but opposite denominators.
  • This structure allows us to simplify using a difference of squares identity: \( (a-b)(a+b) = a^2 - b^2 \).

Applying this formula, the general term simplifies to \( \frac{2}{n-1} \). This simplification reveals that our unfamiliar series can be transformed into a format resembling the harmonic series, making further analysis more straightforward.
Divergence Analysis
Divergence analysis is an essential step in understanding whether a series sums to a finite value or grows indefinitely. In this exercise, once the general term is simplified to \( \frac{2}{n-1} \), we reframe the series as:\(2 \sum_{m=1}^{\infty} \frac{1}{m}\)where we made a substitution to align with the common harmonic series structure.
The task now is to determine if this series converges. Given the properties of the harmonic series, we know that it diverges, as its partial sums increase without limit. Therefore, multiplying by a constant, such as 2 in this case, does not alter the fundamental characteristic of infinite growth.
  • If the base harmonic series \( \sum_{m=1}^{\infty} \frac{1}{m} \) diverges, the scaled version \( 2 \sum_{m=1}^{\infty} \frac{1}{m} \) also diverges.
  • Divergence signifies that no finite sum can describe this series as it continues towards infinity.
Remember, recognizing divergence early can save time and effort in analyzing series behaviors.

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Most popular questions from this chapter

In Problems 1-20, an explicit formula for \(a_{n}\) is given. Write the first five terms of \(\left\\{a_{n}\right\\}\), determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\) \(a_{n}=\frac{n \cos (n \pi)}{2 n-1}\)

In Problems 9-28, find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ \frac{x-2}{1^{2}}+\frac{(x-2)^{2}}{2^{2}}+\frac{(x-2)^{3}}{3^{2}}+\frac{(x-2)^{4}}{4^{2}}+\cdots $$

In Problems 9-28, find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ x+2^{2} x^{2}+3^{2} x^{3}+4^{2} x^{4}+\cdots $$

Assuming that \(u_{1}=\sqrt{3}\) and \(u_{n+1}=\sqrt{3+u_{n}}\) determine a convergent sequence, find \(\lim _{n \rightarrow \infty} u_{n}\) to four decimal places.

Let $$ f(x)= \begin{cases}e^{-1 / x^{2}} & \text { if } x \neq 0 \\ 0 & \text { if } x=0\end{cases} $$ (a) Show that \(f^{\prime}(0)=0\) by using the definition of the derivative. (b) Show that \(f^{\prime \prime}(0)=0\). (c) Assuming the known fact that \(f^{(n)}(0)=0\) for all \(n\), find the Maclaurin series for \(f(x)\). (d) Does the Maclaurin series represent \(f(x)\) ? (e) When \(a=0\), the formula in Theorem \(\mathrm{B}\) is called Maclaurin's Formula. What is the remainder in Maclaurin's Formula for \(f(x)\) ? This shows that a Maclaurin series may exist and yet not represent the given function (the remainder does not tend to 0 as \(n \rightarrow \infty)\).

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