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In Exercises \(57 - 82 ,\) use any method to determine whether the series converges or diverges. Give reasons for your answer. $$ \sum _ { n = 1 } ^ { \infty } \frac { 3 ^ { n } } { n ^ { 3 } } $$

Short Answer

Expert verified
The series diverges by the Ratio Test, as the limit \( L = 3 > 1 \).

Step by step solution

01

Identify the Series Type

The given series is \( \sum_{n=1}^{\infty} \frac{3^n}{n^3} \). We need to determine whether it converges or diverges. Notice that the series has a term that can be expressed as the ratio of a geometric-like term (\(3^n\)) and an algebraic term (\(n^3\)). This suggests using the Ratio Test for convergence.
02

Apply the Ratio Test

The Ratio Test involves calculating \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \), where \( a_n = \frac{3^n}{n^3} \). Calculate:\[\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{3^{n+1}}{(n+1)^3} \cdot \frac{n^3}{3^n} \right| = \frac{3 \cdot n^3}{(n+1)^3}\]Simplify:\[\lim_{n \to \infty} \frac{3n^3}{(n+1)^3} = \lim_{n \to \infty} \frac{3}{1 + \frac{3}{n} + \frac{3}{n^2} + \frac{1}{n^3}} = 3\]
03

Interpret the Ratio Test Result

According to the Ratio Test, if \( L < 1 \), the series converges; if \( L > 1 \), it diverges; and if \( L = 1 \), the test is inconclusive. Here, \( L = 3 > 1 \), so the series \( \sum_{n=1}^{\infty} \frac{3^n}{n^3} \) diverges.

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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 widely utilized method to determine the convergence or divergence of an infinite series. It is particularly useful when dealing with series containing exponential functions or factorials. The basic idea is to compare the ratio of successive terms in the series.
To apply the Ratio Test, consider a series \(\sum_{n=1}^{\infty} a_n\). Compute the limit \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\). Here are the scenarios:
  • If \(L < 1\), the series converges.
  • If \(L > 1\), the series diverges.
  • If \(L = 1\), the test is inconclusive, and other methods must be used.
Applying the test involves simplification after substituting the expressions for \(a_n\) and \(a_{n+1}\). It's crucial to simplify carefully to ensure an accurate calculation of \(L\). The key takeaway here is that the Ratio Test provides a decisive tool for analyzing series, especially when dealing with exponential growth terms.
Geometric Series
A geometric series is one where each term is a constant multiple of the previous term. In other words, the ratio between consecutive terms remains constant. Such series are expressed in the format \(a, ar, ar^2, ar^3, \ldots\), where \(a\) is the first term and \(r\) is the common ratio.
Geometric series are relatively straightforward to analyze in terms of convergence:
  • If \(|r| < 1\), the series converges to the sum \(- \frac{a}{1 - r}\).
  • If \(|r| \geq 1\), the series diverges.
For our problem, though the term \(3^n\) resembles a geometric sequence, it pairs with an algebraic term \(n^3\), preventing the series from being a pure geometric series. Thus, while directly applying geometric series principles isn't possible here, understanding these basics helps frame the mathematical background.
Infinite Series
An infinite series is essentially a sum of an infinite sequence of terms. These series can approach a finite sum – in this case, they converge – or they can grow without bound, thereby diverging.
The challenge of infinite series is determining whether they converge or diverge, which often requires applying a variety of tests and techniques. Some of these include:
  • Limit comparison test.
  • Alternating series test.
  • Ratio and Root tests.
  • Direct comparison test.
Infinite series play a crucial role in calculus and analysis, including in solving differential equations and evaluating limits. The convergence or divergence tells us about the behavior of the series' sums as more terms are added. For the series \(\sum_{n=1}^{\infty} \frac{3^n}{n^3}\), careful analysis using the Ratio Test concludes divergence.
Algebraic Term
Algebraic terms in a sequence or series involve powers of a variable, expressed in the form of polynomials. These terms are typically represented by expressions like \(n^k\), where \(k\) is a constant.
Algebraic terms often appear in combination with other series types (e.g., geometric) to create complex series. In the given problem, \(n^3\) is the algebraic term, paired with the geometric-like exponential term \(3^n\). This combination adds complexity by affecting the dominant behavior of the series.
Algebraic terms influence convergence and divergence differently compared to exponential terms. In many cases, algebraic growth is slower than exponential growth, which can heavily influence the outcome when applying convergence tests. For instance, in the Ratio Test, the presence of \(n^3\) in the denominator affects the limit calculation, leading to the series' divergence in this instance.

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Most popular questions from this chapter

Approximation properties of Taylor polynomials Suppose that \(f(x)\) is differentiable on an interval centered at \(x=a\) and that \(g(x)=b_{0}+b_{1}(x-a)+\cdots+b_{n}(x-a)^{n}\) is a polynomial of degree \(n\) with constant coefficients \(b_{0}, \ldots, b_{n}\) . Let \(E(x)=\) \(f(x)-g(x) .\) Show that if we impose on \(g\) the conditions i) \(E(a)=0\) ii) $$\lim _{x \rightarrow a} \frac{E(x)}{(x-a)^{n}}=0$$ then $$\begin{array}{r}{g(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots} \\ {+\frac{f^{(n)}(a)}{n !}(x-a)^{n}}\end{array}.$$ Thus, the Taylor polynomial \(P_{n}(x)\) is the only polynomial of degree less than or equal to \(n\) whose error is both zero at \(x=a\) and negligible when compared with \((x-a)^{n}.\)

Replace \(x\) by \(-x\) in the Taylor series for \(\ln (1+x)\) to obtain a series for \(\ln (1-x)\) . Then subtract this from the Taylor series for \(\ln (1+x)\) to show that for \(|x|<1\), \begin{equation} \ln \frac{1+x}{1-x}=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right). \end{equation}

Pythagorean triples A triple of positive integers \(a, b,\) and \(c\) is called a Pythagorean triple if \(a^{2}+b^{2}=c^{2} .\) Let \(a\) be an odd positive integer and let $$ b=\left\lfloor\frac{a^{2}}{2}\right\rfloor \text { and } c=\left\lceil\frac{a^{2}}{2}\right\rceil $$ be, respectively, the integer floor and ceiling for \(a^{2} / 2\) a. Show that \(a^{2}+b^{2}=c^{2} .\) (Hint: Let \(a=2 n+1\) and express \(b\) and \(c\) in terms of \(n .\) b. By direct calculation, or by appealing to the accompanying figure, find $$ \lim _{a \rightarrow \infty} \frac{\left\lfloor\frac{a^{2}}{2}\right\rfloor}{ | \frac{a^{2}}{2} \rceil} $$

How many terms of the convergent series \(\sum_{n=1}^{\infty}\left(1 / n^{1.1}\right)\) should be used to estimate its value with error at most 0.00001\(?\)

Assume that the series \(\sum a_{n} x^{n}\) converges for \(x=4\) and diverges for \(x=7 .\) Answer true \((T),\) false \((F),\) or not enough information given \((\mathrm{N})\) for the following statements about the series. $$ \begin{array}{l}{\text { a. Converges absolutely for } x=-4} \\ {\text { b. Diverges for } x=5} \\ {\text { c. Converges absolutely for } x=-8.5} \\\ {\text { d. Converges absolutely for } x=-8.5} \\ {\text { e. Diverges for } x=8} \\ {\text { f. Diverges absolutely for } x=0} \\ {\text { g. Converges absolutely for } x=-7.1}\end{array} $$

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