Math Problem Statement

consider sequence 1+1/(13)+1/(135)++1/(1357)+...++1/(13...*(2n-1)). Is it bounded above?

Solution

To determine whether the given sequence is bounded above, let's analyze the series step by step.


Sequence Description:

The series is

Sn=1+113+1135++1135(2n1)S_n = 1 + \frac{1}{1 \cdot 3} + \frac{1}{1 \cdot 3 \cdot 5} + \dots + \frac{1}{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)}

The nn-th term of the sequence is

an=1135(2n1)a_n = \frac{1}{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)}


Step 1: Bounding Each Term

The nn-th term can be written as: an=1(2n1)!!a_n = \frac{1}{(2n-1)!!} where (2n1)!!(2n-1)!! is the double factorial, representing the product of all odd integers from 11 to (2n1)(2n-1).

For large nn, the growth of the double factorial is rapid. Specifically, it can be approximated as: (2n1)!!(2n)!2nn!.(2n-1)!! \sim \frac{(2n)!}{2^n \cdot n!}.

This shows that ana_n decreases very quickly, indicating convergence of the series.


Step 2: Sum of the Series

We need to check if the series sum, S=n=1anS = \sum_{n=1}^\infty a_n, converges or diverges.

To compare, note that: an=1135(2n1)<1(2n1)!.a_n = \frac{1}{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)} < \frac{1}{(2n-1)!}.

The series n=11(2n1)!\sum_{n=1}^\infty \frac{1}{(2n-1)!} converges because factorial terms grow extremely fast. Therefore, by the comparison test, the given series also converges.


Step 3: Bounded Above

Since the series converges, its partial sums, SnS_n, approach a finite value SS as nn \to \infty. Therefore, the sequence (Sn)(S_n) is bounded above by this finite value SS.


Conclusion:

The sequence of partial sums, SnS_n, is bounded above because the series converges to a finite value.


Would you like a deeper dive into factorial approximations, convergence tests, or other related concepts?


Related Questions:

  1. How does the factorial grow compared to exponential functions?
  2. What is the difference between single and double factorials?
  3. Can we compute an exact value for this series?
  4. How do we use the ratio test to confirm convergence?
  5. How is the comparison test applied in series convergence?

Tip: For rapidly decreasing series, comparing to simpler series like 1n!\frac{1}{n!} or geometric series often helps establish convergence.

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Math Problem Analysis

Mathematical Concepts

Sequences and Series
Double Factorials
Convergence of Series
Comparison Test

Formulas

a_n = 1 / (1 * 3 * 5 * ... * (2n-1))
(2n-1)!! = Product of odd integers from 1 to (2n-1)
Convergence criteria using Comparison Test

Theorems

Comparison Test for Convergence
Factorial Approximation

Suitable Grade Level

Grades 11-12 and above (Advanced High School or College-level Mathematics)