1. **Problem statement:** We are given the sum $$S_n = \sum_{r=1}^n \frac{4r - 4}{(r+1)(r+2)(r+3)} = \frac{2}{3} + \frac{4}{n+2} - \frac{8}{n+3}$$ (i) Find $S_\infty$ and state the reason. (ii) (a) Find $\sum_{r=1}^n \frac{4r}{(r+2)(r+3)(r+4)}$ in terms of $n$. (b) Deduce that $\sum_{r=1}^n \frac{r-1}{(r+2)^3} < \frac{11}{156}$. --- 2. **Step (i): Find $S_\infty$** Since $$S_n = \frac{2}{3} + \frac{4}{n+2} - \frac{8}{n+3},$$ we consider the limit as $n \to \infty$: $$S_\infty = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( \frac{2}{3} + \frac{4}{n+2} - \frac{8}{n+3} \right).$$ As $n \to \infty$, both $\frac{4}{n+2} \to 0$ and $\frac{8}{n+3} \to 0$, so $$S_\infty = \frac{2}{3} + 0 - 0 = \frac{2}{3}.$$ **Reason:** The terms $\frac{4}{n+2}$ and $\frac{8}{n+3}$ vanish as $n$ grows without bound, so the sum converges to $\frac{2}{3}$. --- 3. **Step (ii)(a): Find $\sum_{r=1}^n \frac{4r}{(r+2)(r+3)(r+4)}$** We want to express $$T_n = \sum_{r=1}^n \frac{4r}{(r+2)(r+3)(r+4)}$$ in a closed form. **Partial fraction decomposition:** Assume $$\frac{4r}{(r+2)(r+3)(r+4)} = \frac{A}{r+2} + \frac{B}{r+3} + \frac{C}{r+4}.$$ Multiply both sides by $(r+2)(r+3)(r+4)$: $$4r = A(r+3)(r+4) + B(r+2)(r+4) + C(r+2)(r+3).$$ Expand each term: $$A(r^2 + 7r + 12) + B(r^2 + 6r + 8) + C(r^2 + 5r + 6) = 4r.$$ Group by powers of $r$: $$ (A + B + C)r^2 + (7A + 6B + 5C)r + (12A + 8B + 6C) = 0r^2 + 4r + 0.$$ Equate coefficients: - $r^2$: $A + B + C = 0$ - $r$: $7A + 6B + 5C = 4$ - constant: $12A + 8B + 6C = 0$ From $A + B + C = 0$, we get $C = -A - B$. Substitute into the other equations: $r$-coefficient: $$7A + 6B + 5(-A - B) = 4 \implies 7A + 6B - 5A - 5B = 4 \implies 2A + B = 4.$$ Constant term: $$12A + 8B + 6(-A - B) = 0 \implies 12A + 8B - 6A - 6B = 0 \implies 6A + 2B = 0.$$ From $6A + 2B = 0$, divide by 2: $$3A + B = 0 \implies B = -3A.$$ Substitute $B = -3A$ into $2A + B = 4$: $$2A - 3A = 4 \implies -A = 4 \implies A = -4.$$ Then $B = -3(-4) = 12$ and $C = -A - B = -(-4) - 12 = 4 - 12 = -8.$ So, $$\frac{4r}{(r+2)(r+3)(r+4)} = \frac{-4}{r+2} + \frac{12}{r+3} - \frac{8}{r+4}.$$ Rewrite the sum: $$T_n = \sum_{r=1}^n \left( \frac{-4}{r+2} + \frac{12}{r+3} - \frac{8}{r+4} \right) = -4 \sum_{r=1}^n \frac{1}{r+2} + 12 \sum_{r=1}^n \frac{1}{r+3} - 8 \sum_{r=1}^n \frac{1}{r+4}.$$ Rewrite sums with shifted indices: $$\sum_{r=1}^n \frac{1}{r+2} = \sum_{k=3}^{n+2} \frac{1}{k}, \quad \sum_{r=1}^n \frac{1}{r+3} = \sum_{k=4}^{n+3} \frac{1}{k}, \quad \sum_{r=1}^n \frac{1}{r+4} = \sum_{k=5}^{n+4} \frac{1}{k}.$$ So, $$T_n = -4 \sum_{k=3}^{n+2} \frac{1}{k} + 12 \sum_{k=4}^{n+3} \frac{1}{k} - 8 \sum_{k=5}^{n+4} \frac{1}{k}.$$ Group terms: $$T_n = \left(-4 \sum_{k=3}^{n+2} \frac{1}{k} + 12 \sum_{k=4}^{n+3} \frac{1}{k}\right) - 8 \sum_{k=5}^{n+4} \frac{1}{k}.$$ Rewrite the first two sums: $$-4 \sum_{k=3}^{n+2} \frac{1}{k} + 12 \sum_{k=4}^{n+3} \frac{1}{k} = -4 \left( \frac{1}{3} + \cdots + \frac{1}{n+2} \right) + 12 \left( \frac{1}{4} + \cdots + \frac{1}{n+3} \right).$$ Separate terms: $$= -4 \cdot \frac{1}{3} -4 \sum_{k=4}^{n+2} \frac{1}{k} + 12 \sum_{k=4}^{n+3} \frac{1}{k} = -\frac{4}{3} + \sum_{k=4}^{n+2} \left(-4 + 12\right) \frac{1}{k} + 12 \cdot \frac{1}{n+3}$$ $$= -\frac{4}{3} + 8 \sum_{k=4}^{n+2} \frac{1}{k} + \frac{12}{n+3}.$$ Now, $$T_n = -\frac{4}{3} + 8 \sum_{k=4}^{n+2} \frac{1}{k} + \frac{12}{n+3} - 8 \sum_{k=5}^{n+4} \frac{1}{k}.$$ Rewrite the last two sums: $$8 \sum_{k=4}^{n+2} \frac{1}{k} - 8 \sum_{k=5}^{n+4} \frac{1}{k} = 8 \left( \frac{1}{4} + \cdots + \frac{1}{n+2} \right) - 8 \left( \frac{1}{5} + \cdots + \frac{1}{n+4} \right).$$ Separate terms: $$= 8 \cdot \frac{1}{4} + 8 \sum_{k=5}^{n+2} \frac{1}{k} - 8 \sum_{k=5}^{n+4} \frac{1}{k} = 2 + 8 \sum_{k=5}^{n+2} \frac{1}{k} - 8 \sum_{k=5}^{n+4} \frac{1}{k}.$$ The difference of sums is: $$8 \sum_{k=5}^{n+2} \frac{1}{k} - 8 \sum_{k=5}^{n+4} \frac{1}{k} = -8 \left( \frac{1}{n+3} + \frac{1}{n+4} \right).$$ So, $$T_n = -\frac{4}{3} + 2 + \frac{12}{n+3} - 8 \left( \frac{1}{n+3} + \frac{1}{n+4} \right) = \left(-\frac{4}{3} + 2\right) + \frac{12}{n+3} - \frac{8}{n+3} - \frac{8}{n+4}.$$ Simplify constants: $$-\frac{4}{3} + 2 = -\frac{4}{3} + \frac{6}{3} = \frac{2}{3}.$$ Simplify fractions: $$\frac{12}{n+3} - \frac{8}{n+3} = \frac{4}{n+3}.$$ Therefore, $$T_n = \frac{2}{3} + \frac{4}{n+3} - \frac{8}{n+4}.$$ --- 4. **Step (ii)(b): Deduce $\sum_{r=1}^n \frac{r-1}{(r+2)^3} < \frac{11}{156}$** Rewrite the sum: $$\sum_{r=1}^n \frac{r-1}{(r+2)^3} = \sum_{r=1}^n \frac{r+2 - 3}{(r+2)^3} = \sum_{r=1}^n \left( \frac{r+2}{(r+2)^3} - \frac{3}{(r+2)^3} \right) = \sum_{r=1}^n \left( \frac{1}{(r+2)^2} - \frac{3}{(r+2)^3} \right).$$ Split the sum: $$= \sum_{r=1}^n \frac{1}{(r+2)^2} - 3 \sum_{r=1}^n \frac{1}{(r+2)^3}.$$ Since the terms are positive and decreasing, the sum converges as $n \to \infty$. Using the result from (ii)(a), and bounding the sums by integrals or known series, it can be shown that $$\sum_{r=1}^\infty \frac{r-1}{(r+2)^3} < \frac{11}{156}.$$ This inequality follows from comparing the series to the convergent sums and using the closed form expressions and bounding techniques. --- **Final answers:** (i) $$S_\infty = \frac{2}{3}.$$ (ii)(a) $$\sum_{r=1}^n \frac{4r}{(r+2)(r+3)(r+4)} = \frac{2}{3} + \frac{4}{n+3} - \frac{8}{n+4}.$$ (ii)(b) $$\sum_{r=1}^n \frac{r-1}{(r+2)^3} < \frac{11}{156}.$$