how that the sequence (cn), where cn = (5n + 8)/(3n − 1) for n ≥ 1 isdecreasing. By bounding it below and quoting an appropriate theorem,deduce that it is convergent. Write down (without further justification)the limit.

To prove that the sequence (cn), where cn = (5n + 8)/(3n - 1) for n ≥ 1 is decreasing, we need to show that cn+1 ≤ cn for all n ≥ 1.

Step 1: Proving the sequence is decreasing Let's calculate cn+1 and cn:

cn+1 = (5(n+1) + 8)/(3(n+1) - 1) = (5n + 13)/(3n + 2) cn = (5n + 8)/(3n - 1)

We need to prove that (5n + 13)/(3n + 2) ≤ (5n + 8)/(3n - 1).

Cross multiplying gives us:

(5n + 13)(3n - 1) ≤ (5n + 8)(3n + 2)

Expanding and simplifying gives us:

15n^2 - 5n + 39n - 13 ≤ 15n^2 + 16n + 16

Simplifying further gives us:

-21n ≤ 29

Which is true for all n ≥ 1. Therefore, the sequence is decreasing.

Step 2: Bounding the sequence below The sequence is bounded below by any number less than or equal to the smallest term in the sequence. Since the sequence is decreasing, the smallest term is the limit as n approaches infinity.

Step 3: Quoting an appropriate theorem By the Monotone Convergence Theorem, a sequence that is decreasing and bounded below is convergent.

Step 4: Writing down the limit The limit of the sequence as n approaches infinity can be found by dividing the coefficients of the highest powers of n in the numerator and the denominator.

So, lim (n→∞) (5n + 8)/(3n - 1) = 5/3.