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Use Theorem \(9.11\) to determine the convergence or divergence of the \(p\) -series. \(\sum_{n=1}^{\infty} \frac{1}{n^{1.04}}\)

Short Answer

Expert verified
The p-series \(\sum_{n=1}^{\infty} \frac{1}{n^{1.04}}\) converges.

Step by step solution

01

Identify the p-value in the series

The given series is \(\sum_{n=1}^{\infty} \frac{1}{n^{1.04}}\). Comparing this to the general p-series \(\sum_{n=1}^{\infty} \frac{1}{n^{p}}\), we find that the value of p is 1.04.
02

Use Theorem 9.11 to determine if the series converges or diverges

According to Theorem 9.11, a p-series \(\sum_{n=1}^{\infty} \frac{1}{n^{p}}\) converges if \(p > 1\) and diverges if \(p \leq 1\). From Step 1, we know that in our series, \(p = 1.04 > 1\). Thus, based on Theorem 9.11, our p-series converges.

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Key Concepts

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

Convergence
Convergence in the context of mathematical series refers to the behavior of a series as the number of its terms increases indefinitely. Specifically, a series is said to converge if its sequence of partial sums approaches a finite limit. In simpler terms, adding infinitely many numbers (in the series) results in a specific, well-defined number.

For p-series, convergence is determined by the value of the exponent "p" in the series expression \[\sum_{n=1}^{\infty} \frac{1}{n^p}.\]
A p-series will converge when the value of "p" is greater than 1. This is because a larger "p" makes the terms in the series decrease quickly enough such that they approach zero, enabling the series to sum up to a finite value.
Divergence
Divergence is the opposite of convergence and indicates that a series does not settle to a particular value. In other words, as you add more terms, the partial sums of a divergent series do not approach any fixed number. Instead, they either increase indefinitely or fluctuate without bound.

In the case of a p-series, divergence occurs when the value of "p" is less than or equal to 1. For example, when "p" equals 1, the p-series becomes the harmonic series:\[\sum_{n=1}^{\infty} \frac{1}{n},\]which is well-known to be divergent.
  • Partial sums never settle down to a single finite number.
  • The terms of the series do not shrink fast enough to sum to a limited value.
Theorem
The theorem used in determining the convergence or divergence of a p-series is a crucial part of understanding such series. The specific theorem referenced here is Theorem 9.11, which provides a clear rule:

  • A p-series converges if the exponent "p" is greater than 1.
  • A p-series diverges if the exponent "p" is less than or equal to 1.
This theorem is a handy tool in calculus and analysis because it allows one to quickly determine the behavior of a p-series.
In summary, checking the value of "p" against the threshold of 1 allows us to predict whether the infinite summation will result in a finite quantity or not.
Mathematical Series
Mathematical series are sums of sequences of numbers that follow specific rules. They are foundational concepts in calculus and analysis as they provide ways to handle infinite sequences of terms. One key to understanding a mathematical series is to study their limits and evaluate whether they converge or diverge.

Key characteristics of mathematical series include:
  • Terminology: Each number in the series is called a "term."
  • Partial Sums: The sum of the first n terms gives the nth partial sum.
  • Notation: Represented by the symbol \(\sum\) which denotes summation.
The particular focus here is on p-series, specific series where every term in the sequence is the reciprocal of an integer raised to the power "p." This makes p-series particularly attractive for analysis using elementary convergence criteria like Theorem 9.11.

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Most popular questions from this chapter

The random variable \(n\) represents the number of units of a product sold per day in a store. The probability distribution of \(n\) is given by \(P(n) .\) Find the probability that two units are sold in a given day \([P(2)]\) and show that \(P(1)+P(2)+P(3)+\cdots=1\) $$ P(n)=\frac{1}{2}\left(\frac{1}{2}\right)^{n} $$

Fibonacci Sequence In a study of the progeny of rabbits, Fibonacci (ca. \(1170-\) ca. 1240 ) encountered the sequence now bearing his name. It is defined recursively by \(a_{n+2}=a_{n}+a_{n+1}, \quad\) where \(\quad a_{1}=1\) and \(a_{2}=1\). (a) Write the first 12 terms of the sequence. (b) Write the first 10 terms of the sequence defined by $$ b_{n}=\frac{a_{n+1}}{a_{n}}, \quad n \geq 1 . $$ (c) Using the definition in part (b), show that $$ b_{n}=1+\frac{1}{b_{n-1}} $$ (d) The golden ratio \(\rho\) can be defined by \(\lim _{n \rightarrow \infty} b_{n}=\rho\). Show that \(\rho=1+1 / \rho\) and solve this equation for \(\rho\).

Investigation The interval of convergence of the series \(\sum_{n=0}^{\infty}(3 x)^{n}\) is \(\left(-\frac{1}{3}, \frac{1}{3}\right)\) (a) Find the sum of the series when \(x=\frac{1}{6}\). Use a graphing utility to graph the first six terms of the sequence of partial sums and the horizontal line representing the sum of the series. (b) Repeat part (a) for \(x=-\frac{1}{6}\). (c) Write a short paragraph comparing the rate of convergence of the partial sums with the sum of the series in parts (a) and (b). How do the plots of the partial sums differ as they converge toward the sum of the series? (d) Given any positive real number \(M\), there exists a positive integer \(N\) such that the partial sum \(\sum_{n=0}^{N}\left(3 \cdot \frac{2}{3}\right)^{n}>M\) Use a graphing utility to complete the table. $$ \begin{array}{|l|l|l|l|l|} \hline \boldsymbol{M} & 10 & 100 & 1000 & 10,000 \\ \hline \boldsymbol{N} & & & & \\ \hline \end{array} $$

Explain how to use the series \(g(x)=e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n !}\) to find the series for each function. Do not find the series. (a) \(f(x)=e^{-x}\) (b) \(f(x)=e^{3 x}\) (c) \(f(x)=x e^{x}\) (d) \(f(x)=e^{2 x}+e^{-2 x}\)

show that the function represented by the power series is a solution of the differential equation. $$ y=\sum_{n=0}^{\infty} \frac{x^{2 n}}{2^{n} n !}, \quad y^{\prime \prime}-x y^{\prime}-y=0 $$

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