1. **Stating the problem:** We have a sequence defined by $$u_n = 1\cdot n + 2\cdot (n-1) + \cdots + (n-1)\cdot 2 + n\cdot 1$$ We want to prove that $$u_n = \frac{1}{6} n (n+1)(n+2)$$ Then, for the sequence $v_n$ defined by $$\frac{1}{u_n} = v_n - v_{n+1}$$ we want to find $v_n$, prove $$\sum_{r=1}^n \frac{1}{u_r} = \frac{3}{2} - \frac{3}{(n+1)(n+2)}$$ and finally evaluate $$\sum_{n=1}^\infty \frac{1}{u_n}$$ 2. **Proving the formula for $u_n$:** Rewrite $u_n$ as $$u_n = \sum_{k=1}^n k (n+1 - k)$$ Expand the term inside the sum: $$k (n+1 - k) = k(n+1) - k^2$$ So, $$u_n = \sum_{k=1}^n [k(n+1) - k^2] = (n+1) \sum_{k=1}^n k - \sum_{k=1}^n k^2$$ Use formulas for sums: $$\sum_{k=1}^n k = \frac{n(n+1)}{2}$$ $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$ Substitute: $$u_n = (n+1) \cdot \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}$$ Factor out $n(n+1)$: $$u_n = n(n+1) \left( \frac{n+1}{2} - \frac{2n+1}{6} \right)$$ Simplify inside parentheses: $$\frac{3(n+1) - (2n+1)}{6} = \frac{3n + 3 - 2n - 1}{6} = \frac{n + 2}{6}$$ Therefore, $$u_n = \frac{1}{6} n (n+1)(n+2)$$ 3. **Finding $v_n$ such that $\frac{1}{u_n} = v_n - v_{n+1}$:** We have $$\frac{1}{u_n} = \frac{6}{n(n+1)(n+2)}$$ Try to express this as a telescoping difference: Assume $$v_n = \frac{A}{n(n+1)}$$ Then, $$v_n - v_{n+1} = \frac{A}{n(n+1)} - \frac{A}{(n+1)(n+2)} = A \left( \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right)$$ Find common denominator and simplify: $$= A \frac{(n+2) - n}{n(n+1)(n+2)} = A \frac{2}{n(n+1)(n+2)}$$ We want this to equal $\frac{6}{n(n+1)(n+2)}$, so $$2A = 6 \implies A = 3$$ Hence, $$v_n = \frac{3}{n(n+1)}$$ 4. **Proving the sum formula:** Sum from $r=1$ to $n$: $$\sum_{r=1}^n \frac{1}{u_r} = \sum_{r=1}^n (v_r - v_{r+1}) = v_1 - v_{n+1}$$ Calculate: $$v_1 = \frac{3}{1 \cdot 2} = \frac{3}{2}$$ $$v_{n+1} = \frac{3}{(n+1)(n+2)}$$ So, $$\sum_{r=1}^n \frac{1}{u_r} = \frac{3}{2} - \frac{3}{(n+1)(n+2)}$$ 5. **Evaluating the infinite sum:** Take limit as $n \to \infty$: $$\lim_{n \to \infty} \sum_{r=1}^n \frac{1}{u_r} = \lim_{n \to \infty} \left( \frac{3}{2} - \frac{3}{(n+1)(n+2)} \right) = \frac{3}{2} - 0 = \frac{3}{2}$$ **Final answers:** $$u_n = \frac{1}{6} n (n+1)(n+2)$$ $$v_n = \frac{3}{n(n+1)}$$ $$\sum_{r=1}^n \frac{1}{u_r} = \frac{3}{2} - \frac{3}{(n+1)(n+2)}$$ $$\sum_{n=1}^\infty \frac{1}{u_n} = \frac{3}{2}$$