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Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty} \frac{n^{2} 3^{n}}{3 \cdot 5 \cdot 7 \cdots(2 n+1)}$$

Short Answer

Expert verified
The series converges by the Ratio Test.

Step by step solution

01

Analyze the General Term

The general term of the series is \( a_n = \frac{n^2 \cdot 3^n}{3 \cdot 5 \cdot 7 \cdots (2n+1)} \). It resembles a ratio involving factorial-like growth in the denominator. This suggests that considering the ratio or root test might be beneficial for analysis.
02

Apply the Ratio Test

The Ratio Test is often useful for series with factorial and exponential growth. The test involves examining \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Consider:\[ a_{n+1} = \frac{(n+1)^2 \cdot 3^{n+1}}{3 \cdot 5 \cdot 7 \cdots (2(n+1)+1)} \]The ratio is:\[ \frac{a_{n+1}}{a_n} = \frac{(n+1)^2 \cdot 3^{n+1}}{3 \cdot 5 \cdot 7 \cdots (2n+3)} \times \frac{3 \cdot 5 \cdot 7 \cdots (2n+1)}{n^2 \cdot 3^n} \]Simplifying, it becomes:\[ \frac{a_{n+1}}{a_n} = \frac{(n+1)^2 \cdot 3}{n^2 (2n+3)} \]
03

Simplify the Ratio and Take the Limit

Simplify:\[ \frac{(n+1)^2}{n^2} = 1 + \frac{2}{n} + \frac{1}{n^2} \]The limit becomes:\[ \lim_{n \to \infty} \left( \frac{(n+1)^2 \cdot 3}{n^2 (2n+3)} \right) = \lim_{n \to \infty} \left(3 \cdot \frac{(n+1)^2}{n^2} \cdot \frac{1}{2n+3}\right)\]Breaking it down:- \( \lim_{n \to \infty} \frac{(n+1)^2}{n^2} = 1 \)- \( \lim_{n \to \infty} \frac{1}{2n+3} = 0 \)Therefore, the limiting expression overall evaluates to:\[ 3 \times 1 \times 0 = 0 \]
04

Interpret the Limit Result

According to the Ratio Test, if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \), then the series is absolutely convergent. Here, the limit we found is 0, which is less than 1. Therefore, the given series \( \sum_{n=1}^{\infty} \frac{n^2 3^n}{3 \cdot 5 \cdot 7 \cdots(2n+1)} \) 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 to determine the convergence or divergence of series, especially those with factorial and exponential components. This test involves finding the limit
  • \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]
Here, we examine the absolute value of the ratio of successive terms. If this limit is less than 1, the series converges absolutely. On the other hand, if the limit is greater than 1, the series diverges. If it equals 1, the test is inconclusive.

In the original exercise, for the series \( \sum_{n=1}^{\infty} \frac{n^2 \, 3^n}{3 \cdot 5 \cdot 7 \cdots(2n+1)} \), the Ratio Test was applied effectively. After simplifying, the limit was found to be 0, which is less than 1, proving absolute convergence.
Factorial Growth
Factorial growth often occurs in series involving products or sequences that extend to infinity, like \( 3 \cdot 5 \cdot 7 \cdots (2n+1) \). Factorials grow extremely fast, usually outpacing polynomial terms in the numerator, thus often making them a key determinant in convergence questions.

This concept becomes crucial when applying tests like the Ratio Test on series. In the original problem, despite the polynomial \( n^2 \) and exponential \( 3^n \) terms in the numerator, the factorial-like growth in the denominator plays a decisive role. The rapid escalation of these terms in the denominator helped produce a limit that is less than one, confirming the series' convergence.
Limit Evaluation
Limit evaluation is a necessary step in applying the Ratio Test. It involves simplifying expressions to make the computation of limits feasible. Given the series in the exercise, simplify:
  • \[ \frac{(n+1)^2}{n^2} = 1 + \frac{2}{n} + \frac{1}{n^2} \]
When \( n \to \infty \), terms like \( \frac{2}{n} \) and \( \frac{1}{n^2} \) shrink to zero, meaning the overall expression converges to a manageable value.

This simplification enables us to evaluate:
  • \[ \lim_{n \to \infty} \left(3 \cdot \frac{(n+1)^2}{n^2} \cdot \frac{1}{2n+3}\right) \]
Breaking down each component in terms of limits leads to understanding how and why the series converges based on the Ratio Test.
Infinite Series
Infinite series are sums involving an infinite number of terms. They can converge, resulting in a finite sum, or diverge, tending towards infinity or no particular limit. Understanding whether a given infinite series converges or diverges is crucial in numerous mathematical and applied contexts.

The original exercise dealt with a complex infinite series expressed as \( \sum_{n=1}^{\infty} \frac{n^2 \, 3^n}{3 \cdot 5 \cdot 7 \cdots (2n+1)} \). By applying the Ratio Test, we determined the series converges absolutely. This type of analysis is useful across many fields, indicating a predictable and stable behavior of functions that can be represented as series.

Ultimately, mastering infinite series is critical for handling sequences and series in calculus and understanding their broader applications in real-world scenarios.

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Most popular questions from this chapter

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty} \frac{(-3)^{n}}{n^{3} 2^{n}}$$

Show that the sum of the first \(2 n\) terms of the series $$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\cdots$$ is the same as the sum of the first \(n\) terms of the series $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\frac{1}{5 \cdot 6}+\cdots$$ Do these series converge? What is the sum of the first \(2 n+1\) terms of the first series? If the series converge, what is their sum?

Assume that the series \(\sum a_{n}(x-2)^{n}\) converges for \(x=-1\) and diverges for \(x=6 .\) Answer true (T), false (F), or not enough information given (N) for the following statements about the series. a. Converges absolutely for \(x=1\) b. Diverges for \(x=-6\) c. Diverges for \(x=2\) d. Converges for \(x=0\) e. Converges absolutely for \(x=5\) f. Diverges for \(x=4.9\) g. Diverges for \(x=5.1\) h. Converges absolutely for \(x=4\)

Show that neither the Ratio Test nor the Root Test provides information about the convergence of $$\sum_{n=2}^{\infty} \frac{1}{(\ln n)^{p}} \quad(p \text { constant })$$

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 \(b_{0} \ldots . . b_{m-}\) Let \(E(x)=\) \(f(x)-g(x) .\) Show that if we impose on \(g\) the conditions i) \(E(a)=0 \quad\) The approximation crror is acro at \(x=a\) ii) \(\lim _{x \rightarrow a} \frac{E(x)}{(x-a)^{n}}=0, \quad \begin{array}{l}\text { The error is negligible when compared to } \\\ (x-a)^{n}\end{array}\) then $$g(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots$$ $$+\frac{f^{[n |}(a)}{n !}(x-a)^{n}$$ 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 \((x-a)^{n}\)

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