Math Problem Statement

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solve problem 4 h

Solution

The relevant content from the PDF includes Problem 4, which involves determining the convergence or divergence of several series. Here’s a summary of the problem components:

Problem 4

Determine whether the following series converge or diverge, and state why. If the series converges, find its sum.

(a) n=111+3n5\sum_{n=1}^{\infty} \frac{1}{1 + 3n^5}

(b) n=12n1+4n+3n2\sum_{n=1}^{\infty} \frac{2n}{1 + 4n + 3n^2}

(c) n=11n(n+1)(n+3)\sum_{n=1}^{\infty} \frac{1}{n(n + 1)(n + 3)}

(d) n=1ensin(2n)\sum_{n=1}^{\infty} e^{-n} \sin\left(\frac{2}{n}\right) (Hint: limnsin(2n)2/n=1\lim_{n \to \infty} \frac{\sin\left(\frac{2}{n}\right)}{2/n} = 1)

(e) n=15n2n+2n\sum_{n=1}^{\infty} \frac{5^n}{2^n + 2n}

(f) n=23nn2+9\sum_{n=2}^{\infty} \frac{3^n}{n^2 + 9}

(g) 1+0+39+027+1 + 0 + \frac{3}{9} + \frac{0}{27} + \ldots

(h) n=1(13n+29n)\sum_{n=1}^{\infty} \left(\frac{1}{3^n} + \frac{2}{9^n}\right)

Let's analyze each part for convergence or divergence.

(a) n=111+3n5\sum_{n=1}^{\infty} \frac{1}{1 + 3n^5}

  • Comparison Test: Compare with 13n5\frac{1}{3n^5}. Since 1n5\sum \frac{1}{n^5} converges, so does this series.

(b) n=12n1+4n+3n2\sum_{n=1}^{\infty} \frac{2n}{1 + 4n + 3n^2}

  • Asymptotic Behavior: As nn \to \infty, 2n3n2=23n\frac{2n}{3n^2} = \frac{2}{3n}. Since 1n\sum \frac{1}{n} diverges, this series diverges.

(c) n=11n(n+1)(n+3)\sum_{n=1}^{\infty} \frac{1}{n(n + 1)(n + 3)}

  • Partial Fraction Decomposition: 1n(n+1)(n+3)=An+Bn+1+Cn+3\frac{1}{n(n + 1)(n + 3)} = \frac{A}{n} + \frac{B}{n + 1} + \frac{C}{n + 3}. This leads to a telescoping series, which converges.

(d) n=1ensin(2n)\sum_{n=1}^{\infty} e^{-n} \sin\left(\frac{2}{n}\right)

  • Limit Comparison Test: limnensin(2n)en2n\lim_{n \to \infty} e^{-n} \sin\left(\frac{2}{n}\right) \approx e^{-n} \cdot \frac{2}{n}. Since en/n\sum e^{-n}/n converges, this series converges.

(e) n=15n2n+2n\sum_{n=1}^{\infty} \frac{5^n}{2^n + 2n}

  • Ratio Test: The dominant term 5n/2n5^n/2^n leads to divergence because the series behaves like (2.5)n\sum (2.5)^n.

(f) n=23nn2+9\sum_{n=2}^{\infty} \frac{3^n}{n^2 + 9}

  • Comparison Test: Since n2n^2 dominates, compare with 3nn2\sum \frac{3^n}{n^2}, which diverges.

(g) 1+0+39+027+1 + 0 + \frac{3}{9} + \frac{0}{27} + \ldots

  • Geometric Series: The series has terms that converge to zero and effectively diverge due to the nature of the non-zero terms.

(h) n=1(13n+29n)\sum_{n=1}^{\infty} \left(\frac{1}{3^n} + \frac{2}{9^n}\right)

  • Geometric Series: This is a sum of two converging geometric series, both converging to 1/311/3+2/911/9\frac{1/3}{1-1/3} + \frac{2/9}{1-1/9}.

Summary of Results

  • (a) Converges
  • (b) Diverges
  • (c) Converges
  • (d) Converges
  • (e) Diverges
  • (f) Diverges
  • (g) Diverges
  • (h) Converges

Would you like detailed solutions or explanations for any specific part? Here are some related questions:

  1. How do you apply the Comparison Test in detail?
  2. What is partial fraction decomposition, and how is it used in series?
  3. Can you explain the Ratio Test with examples?
  4. What conditions determine the convergence of geometric series?
  5. How do the behavior of terms in a series affect convergence?

Tip: Always check the behavior of the terms as nn approaches infinity to decide on convergence.

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Math Problem Analysis

Mathematical Concepts

Series and Sequences
Convergence Tests
Geometric Series

Formulas

Comparison Test
Limit Comparison Test
Ratio Test
Geometric Series Sum Formula

Theorems

Comparison Test
Limit Comparison Test
Ratio Test

Suitable Grade Level

Grades 11-12