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(a) Determine whether the sequence is increasing, decreasing, or not monotonic. (Assume that nn begins with 11.) an=3n44n+3a_n = \frac{3n - 4}{4n + 3}

  • increasing
  • decreasing
  • not monotonic

(b) Is the sequence bounded?

  • bounded
  • not bounded

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Question :

(a) determine whether the sequence is increasing, decreasing, or not monotonic. (assume that nn begins with 11.) an=3n44n+3a_n = \frac{3n - 4}{4n + 3}

  • increasing
  • decreasing
  • not monotonic

(b) is the sequence bounded?

  • bounded
  • not bounded

(a) determine whether the sequence is increasing, decreasing, or not monoton | Doubtlet.com

Solution:

Neetesh Kumar

Neetesh Kumar | December 31, 2024

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This is the solution to Math 1c
Assignment: 11.1 Question Number 37
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Step-by-step solution:

Part (a): Determine monotonicity:

Step 1: Compute an+1a_{n+1} and compare it to ana_n:

The sequence is an=3n44n+3a_n = \frac{3n - 4}{4n + 3}.

To determine if it is increasing or decreasing, we analyze the difference an+1ana_{n+1} - a_n.

Step 2: Write an+1a_{n+1}:

Substitute n+1n+1 into the formula for ana_n:

an+1=3(n+1)44(n+1)+3=3n+344n+4+3=3n14n+7a_{n+1} = \frac{3(n+1) - 4}{4(n+1) + 3} = \frac{3n + 3 - 4}{4n + 4 + 3} = \frac{3n - 1}{4n + 7}

Step 3: Analyze an+1ana_{n+1} - a_n:

Compute the difference:

an+1an=3n14n+73n44n+3a_{n+1} - a_n = \frac{3n - 1}{4n + 7} - \frac{3n - 4}{4n + 3}

Combine under a common denominator:

an+1an=(3n1)(4n+3)(3n4)(4n+7)(4n+7)(4n+3)a_{n+1} - a_n = \frac{(3n - 1)(4n + 3) - (3n - 4)(4n + 7)}{(4n + 7)(4n + 3)}

Expand the numerator:

(3n1)(4n+3)=12n2+9n4n3=12n2+5n3(3n - 1)(4n + 3) = 12n^2 + 9n - 4n - 3 = 12n^2 + 5n - 3

(3n4)(4n+7)=12n2+21n16n28=12n2+5n28(3n - 4)(4n + 7) = 12n^2 + 21n - 16n - 28 = 12n^2 + 5n - 28

Subtract the two terms:

an+1an=(12n2+5n3)(12n2+5n28)(4n+7)(4n+3)a_{n+1} - a_n = \frac{(12n^2 + 5n - 3) - (12n^2 + 5n - 28)}{(4n + 7)(4n + 3)}

Simplify the numerator:

12n2+5n312n25n+28=2512n^2 + 5n - 3 - 12n^2 - 5n + 28 = 25

Thus:

an+1an=25(4n+7)(4n+3)a_{n+1} - a_n = \frac{25}{(4n + 7)(4n + 3)}

Step 4: Sign of an+1ana_{n+1} - a_n:

The denominator (4n+7)(4n+3)>0(4n + 7)(4n + 3) > 0 for all n1n \geq 1, and the numerator 25>025 > 0.

Therefore:

an+1an>0a_{n+1} - a_n > 0

This shows that the sequence is increasing.

Part (b): Determine boundedness:

Step 1: Check for upper and lower bounds:

The sequence is:

an=3n44n+3a_n = \frac{3n - 4}{4n + 3}

For large nn, divide numerator and denominator by nn:

an=34n4+3n34as na_n = \frac{3 - \frac{4}{n}}{4 + \frac{3}{n}} \to \frac{3}{4} \quad \text{as } n \to \infty

For n=1n = 1, a1=3(1)44(1)+3=17=17a_1 = \frac{3(1) - 4}{4(1) + 3} = \frac{-1}{7} = -\frac{1}{7}. Therefore, the sequence is bounded below by 17-\frac{1}{7} and above by 34\frac{3}{4}.

Step 2: Conclusion:

The sequence is bounded.

Final Answers:

(a) The sequence is increasing\boxed{\text{increasing}}

(b) The sequence is bounded\boxed{\text{bounded}}


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