Sequences & Series Question - None

Question

A sequence of positive terms A_1, A_2, A_3, \ldots, A_n satisfies the relation A_{n+1} = \frac{3(1 + A_n)}{3 + A_n}. Find the least integral value of A_1 for which the sequence is decreasing.

Accepted Answer

Step 1: Given A_{n+1} = \frac{3(1 + A_n)}{3 + A_n}

Step 2: For n = 1, A_2 = \frac{3(1 + A_1)}{3 + A_1}

Step 3: For n = 2, A_3 = \frac{3(1 + A_2)}{3 + A_2}

Step 4: The sequence can be written as: A_1, \frac{3(1 + A_1)}{3 + A_1}, \frac{3(1 + A_2)}{3 + A_2}, \ldots

Step 5: Given A_1 > 0 and the sequence is decreasing, we need A_1 > A_2

Step 6: A_1 > \frac{3(1 + A_1)}{3 + A_1}

Step 7: A_1(3 + A_1) > 3(1 + A_1)

Step 8: 3A_1 + A_1^2 > 3 + 3A_1

Step 9: A_1^2 > 3

Step 10: A_1 > \sqrt{3} \approx 1.732

∴ The least integral value of A_1 is 2.

Question:

Step-by-Step Solution

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