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Chapter 10: Problem 21

In Exercises \(17-46,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$ \sum_{n=1}^{\infty} \frac{n^{10}}{10^{n}} $$

Short Answer

Expert verified
The series \( \sum_{n=1}^{\infty} \frac{n^{10}}{10^n} \) converges by the Ratio Test.

Step by step solution

01

Identify the Series Type

The given series is \( \sum_{n=1}^{\infty} \frac{n^{10}}{10^n} \). This is a series with terms of the form \( \frac{f(n)}{r^n} \), where \( f(n) = n^{10} \). It resembles a form suitable for the Ratio Test due to the exponential \( 10^n \) in the denominator.
02

Apply the Ratio Test

The Ratio Test states that for a series \( \sum a_n \), if \( \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = L \) and \( L < 1 \), the series converges. Here, \( a_n = \frac{n^{10}}{10^n} \).Find \( a_{n+1} = \frac{(n+1)^{10}}{10^{n+1}} \) and compute the ratio:\[\left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)^{10}}{10^{n+1}} \cdot \frac{10^n}{n^{10}} = \frac{(n+1)^{10}}{10 \cdot n^{10}}\]
03

Simplify the Ratio

Simplify the ratio derived in Step 2:\[\frac{(n+1)^{10}}{10 \cdot n^{10}} = \frac{1}{10} \left( \frac{n+1}{n} \right)^{10} = \frac{1}{10} \left( 1 + \frac{1}{n} \right)^{10}\]
04

Evaluate the Limit

Evaluate the limit:\[L = \lim_{n\to\infty} \frac{1}{10} \left( 1 + \frac{1}{n} \right)^{10} = \frac{1}{10} \times 1^{10} = \frac{1}{10}\]Since \( L = \frac{1}{10} < 1 \), by the Ratio Test, the series converges.

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Key Concepts

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

Ratio Test
The Ratio Test is a powerful tool used to determine the convergence of a series. If you have a series like \( \sum a_n \), the Ratio Test suggests you examine the limit of the absolute value of the ratio of consecutive terms. If
  • \( L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \), the series converges.
  • \( L > 1 \) or the limit is infinite, the series diverges.
  • \( L = 1 \), the test is inconclusive.
In our example, the series \( \sum_{n=1}^{\infty} \frac{n^{10}}{10^n} \) fits perfectly for the Ratio Test due to the exponential term \( 10^n \) in the denominator. We derived the limit \( L = \frac{1}{10} \), showing the series converges.
Infinite Series
An infinite series is a sum of an infinite sequence of terms. Each term is defined by a specific formula involving an index like \( n \). Series often take forms where terms shrink exponentially as \( n \) increases.

Infinite series can sum to a finite value, converge, or can grow indefinitely, diverge.
Convergence and divergence depend on how the terms behave as \( n \) approaches infinity. Math tools like the Ratio Test help us identify the behavior. In our case, analyzing infinite series, we're particularly interested in the summation from \( n=1 \) to infinity.
Exponential Terms
Exponential terms include expressions like \( 10^n \) in the denominator. These terms grow extremely fast as \( n \) increases.

This rapid growth impacts the series significantly, often leading the terms towards zero when evaluated.
For instance, \( \frac{n^{10}}{10^n} \) rapidly becomes smaller as \( n \) increases, because \( 10^n \) grows more quickly than \( n^{10} \).

When you see these in a series, consider the Ratio Test, which examines the effect of such fast-growing terms on the convergence of a series.
Limit Evaluation
Limit evaluation is crucial in determining convergence. This process involves calculating the limit of the ratio of consecutive terms of a series as \( n \) approaches infinity.

In simple terms, you're checking how the series terms behave as the series grows. The outcome guides us on the convergence behavior. For the series \( \sum_{n=1}^{\infty} \frac{n^{10}}{10^n} \), we find the limit of the ratio of successive terms.
  • We simplified the ratio, using limits to reveal that \( \lim_{n\to\infty} \left( \frac{1}{10} \right)^{10} = \frac{1}{10} \).
This confirms convergence as it is less than one.

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Most popular questions from this chapter

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L\) ? b. If the sequence converges, find an integer \(N\) such that \(\quad\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{1}=1, \quad a_{n+1}=a_{n}+\frac{1}{5^{n}} $$

Prove that $$ \lim _{n \rightarrow \infty} \sqrt[n]{n}=1 $$

Logarithmic \(p\) -series $$ \begin{array}{c}{\text { a. Show that the improper integral }} \\\ {\int_{2}^{\infty} \frac{d x}{x(\ln x)^{p}} \quad(p \text { a positive constant })} \\ {\text { converges if and only if } p>1} \\ {\text { b. What implications does the fact in part (a) have for the con- }} \\\ {\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}} ?} \\ {\text { Give reasons for vour answer. }}\end{array} $$

By multiplying the Taylor series for \(e^{x}\) and \(\sin x,\) find the terms through \(x^{5}\) of the Taylor series for \(e^{x} \sin x .\) This series is the imaginary part of the series for \begin{equation}e^{x} \cdot e^{i x}=e^{(1+i) x}.\end{equation} Use this fact to check your answer. For what values of \(x\) should the series for \(e^{x} \sin x\) converge?

When \(a\) and \(b\) are real, we define \(e^{(a+i b) x}\) with the equation \begin{equation}e^{(a+i b) x}=e^{a x} \cdot e^{i b x}=e^{a x}(\cos b x+i \sin b x).\end{equation} Differentiate the right-hand side of this equation to show that \begin{equation}\frac{d}{d x} e^{(a+i b) x}=(a+i b) e^{(a+i b) x}.\end{equation} Thus the familiar rule \((d / d x) e^{k x}=k e^{k x}\) holds for \(k\) complex as well as real.
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