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Chapter 10: Problem 51

Which of the series in Exercises \(17-56\) converge, and which diverge? Use any method, and give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{1}{n \sqrt[n]{n}} $$

Short Answer

Expert verified
The series converges by the Root Test.

Step by step solution

01

Identify the Series Type

The given series is \( \sum_{n=1}^{\infty} \frac{1}{n \sqrt[n]{n}} \). This series involves a fraction with \( n \) in the denominator raised to two different powers. We need to determine convergence using a suitable test.
02

Apply Root Test

The Root Test is useful for series where terms involve roots. For the Root Test, compute \( \lim_{n \to \infty} \sqrt[n]{a_n} \), where \( a_n = \frac{1}{n\sqrt[n]{n}} \).
03

Simplify and Calculate Limit

First, find the expression for the nth root: \( \sqrt[n]{a_n} = \sqrt[n]{\frac{1}{n\sqrt[n]{n}}} = \left(\frac{1}{n} \right)^{1/n} \). Then \( \lim_{n \to \infty} \left(\frac{1}{n} \right)^{1/n} \cdot \frac{1}{n^{1/n}} \).
04

Evaluate the Limit

The term \( \left(\frac{1}{n} \right)^{1/n} \) simplifies to \( n^{-1/n} \), which as \( n \) approaches infinity, goes to \( 1^0 = 1 \). Similarly, \( n^{1/n} \) also approaches \( 1 \). Thus, the overall limit is 0.
05

Conclusion with Root Test

Since \( \lim_{n \to \infty} \sqrt[n]{a_n} = 0 < 1 \), the Root Test confirms the series converges.

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Key Concepts

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

Root Test
The Root Test is a powerful tool in determining the convergence or divergence of an infinite series. It's especially useful when the terms of the series involve roots or exponential expressions. The idea is to take the nth root of each term in the series.

The mathematical formula for the Root Test is:
  • Compute \( L = \lim_{n \to \infty} \sqrt[n]{|a_n|} \)
The outcome is interpreted as follows:
  • If \( L < 1 \), the series converges absolutely.
  • If \( L > 1 \), the series diverges.
  • If \( L = 1 \), the test is inconclusive.
In the given series \( \sum_{n=1}^{\infty} \frac{1}{n \sqrt[n]{n}} \), the Root Test is applied by computing the limit of the nth root of the absolute value of the terms, leading to the conclusion that the series converges.
Limit Evaluation
Limit evaluation is a critical component of many convergence tests, including the Root Test. It involves calculating the limit of a particular expression as the variable (usually \( n \)) approaches infinity.

For the Root Test, this means evaluating the limit of the nth root of the series terms. In our series example,
\[ \lim_{n \to \infty} \sqrt[n]{\left(\frac{1}{n}\sqrt[n]{n}\right)} \]
This simplifies step by step:
  • \( \sqrt[n]{\frac{1}{n \sqrt[n]{n}}} = \left(\frac{1}{n}\right)^{1/n} \)
  • The term \( n^{1/n} \rightarrow 1 \) as \( n \rightarrow \infty \)
  • The expression \( \left(\frac{1}{n}\right)^{1/n} \rightarrow 1 \) as \( n \rightarrow \infty \)
Thus, the limit simplifies to 0, which confirms convergence via the Root Test.
Infinite Series
An infinite series is the sum of an infinite sequence of numbers, often denoted by the summation symbol \( \sum \). Understanding whether such a series converges to a finite number or diverges (grows indefinitely) is crucial in analysis.

Infinite series take forms like:
  • Arithmetic series
  • Geometric series
  • Series involving factorials or polynomial expressions
  • Series that require special tests (e.g., Root Test, Ratio Test)
In the series \( \sum_{n=1}^{\infty} \frac{1}{n \sqrt[n]{n}} \), we see a unique combination of polynomial and root elements, making the Root Test ideal for evaluating its convergence.
Convergence and Divergence
Convergence and divergence describe whether an infinite series approaches a specific value or not. A series converges if its sequence of partial sums tends to a limit. Conversely, if the sums grow indefinitely or oscillate, the series diverges.

Some tips to assess series convergence:
  • Recognize common series forms, such as geometric or harmonic series.
  • Use convergence tests like the Root Test, Ratio Test, or the Integral Test.
  • Be mindful that a convergent series doesn't always imply convergent terms (except for absolutely convergent series).
In our example, since the Root Test result is less than 1, the series is determined to converge, reaching a finite limit as \( n \to \infty \). Understanding these principles allows for informed and accurate evaluations of infinite series.

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Most popular questions from this chapter

Linearizations at inflection points Show that if the graph of a twice- differentiable function \(f(x)\) has an inflection point at \(x=a\) , then the linearization of \(f\) at \(x=a\) is also the quadratic approximation of \(f\) at \(x=a\) . This explains why tangent lines fit so well at inflection points.

\begin{equation} \begin{array}{l}{\text { Estimate the error if } P_{4}(x)=1+x+\left(x^{2} / 2\right)+\left(x^{3} / 6\right)+\left(x^{4} / 24\right)} \\ {\text { is used to estimate the value of } e^{x} \text { at } x=1 / 2}\end{array} \end{equation}

Estimating Pi About how many terms of the Taylor series for \(\tan ^{-1} x\) would you have to use to evaluate each term on the right- hand side of the equation \begin{equation} \pi=48 \tan ^{-1} \frac{1}{18}+32 \tan ^{-1} \frac{1}{57}-20 \tan ^{-1} \frac{1}{239} \end{equation} with an error of magnitude less than \(10^{-6}\) ? In contrast, the convergence of \(\sum_{n=1}^{\infty}\left(1 / n^{2}\right)\) to \(\pi^{2} / 6\) is so slow that even 50 terms will not yield two-place accuracy.

Which of the series in Exercises 13 46 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{8 \tan ^{-1} n}{1+n^{2}} $$

Which of the series in Exercises 13 46 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=2}^{\infty} \frac{\sqrt{n+2}-\sqrt{n+1}}{\sqrt{n+1} \sqrt{n+2}} $$
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