Chapter 11.1, Problem 37E

### Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643

Chapter
Section

### Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643
Textbook Problem

# Determine whether the sequence converges or diverges. If it converges, find the limit.37. { ( 2 n − 1 ) ! ( 2 n + 1 ) ! }

To determine

Whether the sequence converges or diverges and obtain the limit if the sequence converges.

Explanation

Given:

Consider the given sequence as an={(2n1)!(2n+1)!} .

Definition used:

If an is a sequence and limnan exists, then the sequence an is said to be converges; otherwise it diverges.

Laws of limits for sequences used:

If f(x) and g(x) are two functions, then, limxa[f(x)g(x)]=limxaf(x)limxag(x) and limxag(x)0 .

If f(x) and g(x) are two functions, then, limxa[f(x)+g(x)]=limxaf(x)+limxag(x) .

Calculation:

Let an={(2n1)!(2n+1)!} .

Obtain the limit of the sequence to investigate whether the sequence converges or diverges.

limnan=limn(2n1)!(2n+1)!=limn(2n1)!(2n+1)(2n)(2n1)!=limn1(2n+1)(2n)=limn14n2+2n

Divide numerator and the denominator by the highest power

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