Given that: \( \sum_{n=1}^{\infty} 63 m^{n-1}=\frac{189}{2} \) \( 3.1 \quad \) Solve for \( m \). \( 3.2 \quad \) If it is further given that \( m=\frac{1}{3} \), determine the smallest value of \( n \) such that \( 1 T_{n}<\frac{1}{6561} \)

We start with the given infinite series   Σₙ₌₁∞ 63·m^(n–1) = 189⁄2. Step 1. (Finding m) A geometric series with first term a = 63 and common ratio r = m converges (when |m| < 1) to   S = a⁄(1 – m). Thus, we have   63⁄(1 – m) = 189⁄2. Solve for m:   Multiply both sides by (1 – m):     63 = (189⁄2)(1 – m).   Multiply both sides by 2 to eliminate the fraction:     126 = 189(1 – m).   Divide both sides by 189:     1 – m = 126⁄189 = 2⁄3.   Thus, m = 1 – (2⁄3) = 1⁄3. Step 2. (Finding the Smallest n Given m = 1⁄3) We now wish to determine the smallest integer n such that   1 Tₙ < 1⁄6561. The notation “Tₙ” is taken to mean the remainder (also called the tail) after summing the first n terms of the series. Recall that for a geometric series, the sum to infinity is S∞ = a⁄(1 – m) and the sum of the first n terms is   Sₙ = a(1 – mⁿ)⁄(1 – m). Thus, the remainder (or error) after n terms is   Rₙ = S∞ – Sₙ = a·mⁿ⁄(1 – m). For our series with a = 63 and m = 1⁄3,   Rₙ = 63·(1⁄3)ⁿ⁄(1 – 1⁄3) = 63·(1⁄3)ⁿ⁄(2⁄3) = (63·3)/(2)·(1⁄3)ⁿ = (189⁄2)·(1⁄3)ⁿ. We need Rₙ < 1⁄6561, that is:   (189⁄2)·(1⁄3)ⁿ < 1⁄6561. To solve for n, first multiply both sides by 2⁄189:   (1⁄3)ⁿ < (1⁄6561)·(2⁄189). Notice we can simplify using powers of 3:   • 6561 is a power of 3 since 3⁸ = 6561.   • 189 can be written as 3³·7. Thus, the right-hand side becomes:   (2)⁄(189·6561) = 2⁄((3³·7)·3⁸) = 2⁄(7·3^(3+8)) = 2⁄(7·3¹¹). So the inequality is:   1⁄3ⁿ < 2⁄(7·3¹¹). Taking reciprocals (and remembering that taking reciprocals reverses the inequality, but here it’s simpler to rearrange by multiplying both sides by 3ⁿ): Multiply both sides by 3ⁿ:   1 < (2⁄(7·3¹¹))·3ⁿ. Multiply both sides by (7·3¹¹)⁄2:   (7·3¹¹)⁄2 < 3ⁿ. We can also write this inequality as:   3ⁿ > (7⁄2)·3¹¹. Divide both sides by 3¹¹:   3^(n–11) > 7⁄2     (7⁄2 = 3.5). Now we look for the smallest integer n – 11 such that 3^(n–11) > 3.5. Check:   If n – 11 = 1, then 3¹ = 3 which is not > 3.5.   If n – 11 = 2, then 3² = 9 which is > 3.5. Thus, the smallest possible value is when n – 11 = 2, so   n = 11 + 2 = 13. Answer Summary: 3.1) m = 1⁄3. 3.2) The smallest integer n such that the tail (remainder) is less than 1⁄6561 is n = 13.