1. **Problem statement:** Given the sequence $u_n = \frac{n^2}{2^n}$ for $n > 0$, define $v_n = \frac{u_{n+1}}{u_n}$. We need to find $\lim_{n \to +\infty} v_n$, prove inequalities about $v_n$, find a threshold $N$, and analyze sums $S_n$. 2. **Formula for $v_n$:** $$v_n = \frac{u_{n+1}}{u_n} = \frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}} = \frac{(n+1)^2}{2^{n+1}} \times \frac{2^n}{n^2} = \frac{(n+1)^2}{2 n^2}$$ 3. **a) Find $\lim_{n \to +\infty} v_n$:** $$\lim_{n \to +\infty} v_n = \lim_{n \to +\infty} \frac{(n+1)^2}{2 n^2} = \frac{1}{2} \lim_{n \to +\infty} \left(1 + \frac{1}{n}\right)^2 = \frac{1}{2} \times 1 = \frac{1}{2}$$ 4. **b) Prove $v_n > \frac{1}{2}$ for all $n > 0$:** Since $v_n = \frac{(n+1)^2}{2 n^2} = \frac{1}{2} \left(1 + \frac{1}{n}\right)^2$, and $\left(1 + \frac{1}{n}\right)^2 > 1$ for all $n > 0$, it follows that $$v_n > \frac{1}{2}$$ 5. **c) Find smallest $N$ such that for $n \geq N$, $v_n < \frac{3}{4}$:** Solve $$\frac{(n+1)^2}{2 n^2} < \frac{3}{4} \implies 2 (n+1)^2 < 3 n^2 \implies 2 (n^2 + 2n + 1) < 3 n^2$$ $$2 n^2 + 4 n + 2 < 3 n^2 \implies 0 < 3 n^2 - 2 n^2 - 4 n - 2 = n^2 - 4 n - 2$$ Solve quadratic inequality: $$n^2 - 4 n - 2 > 0$$ Roots: $$n = \frac{4 \pm \sqrt{16 + 8}}{2} = 2 \pm \sqrt{6}$$ Since $n > 0$, inequality holds for $$n > 2 + \sqrt{6} \approx 4.45$$ So smallest natural $N$ is 5. 6. **d) Deduce for $n \geq N$, $u_{n+1} < \frac{3}{4} u_n$:** By definition, $$v_n = \frac{u_{n+1}}{u_n} < \frac{3}{4} \implies u_{n+1} < \frac{3}{4} u_n$$ for all $n \geq 5$. 7. **2a) Prove $u_n \leq \left(\frac{3}{4}\right)^{n-5} u_5$ for $n \geq 5$:** By induction and using $u_{n+1} < \frac{3}{4} u_n$ for $n \geq 5$, we get $$u_n < \left(\frac{3}{4}\right)^{n-5} u_5$$ 8. **2b) Show $S_n = u_5 + u_6 + ... + u_n \leq \left(1 + \frac{3}{4} + \left(\frac{3}{4}\right)^2 + ... + \left(\frac{3}{4}\right)^{n-5}\right) u_5$:** Since each $u_k \leq \left(\frac{3}{4}\right)^{k-5} u_5$, sum is bounded by geometric series: $$S_n \leq u_5 \sum_{k=0}^{n-5} \left(\frac{3}{4}\right)^k$$ 9. **2c) Deduce $S_n \leq 4 u_5$ for $n \geq 5$:** Sum of geometric series with ratio $r=\frac{3}{4}$: $$\sum_{k=0}^m r^k = \frac{1 - r^{m+1}}{1 - r} < \frac{1}{1 - \frac{3}{4}} = 4$$ Thus, $$S_n \leq 4 u_5$$ 10. **2d) Show $(S_n)$ is increasing and convergent:** Since $u_n > 0$, $S_n$ is increasing. Also, $S_n$ is bounded above by $4 u_5$, so by Monotone Convergence Theorem, $(S_n)$ converges. **Final answers:** - $\lim_{n \to +\infty} v_n = \frac{1}{2}$ - $v_n > \frac{1}{2}$ for all $n > 0$ - Smallest $N = 5$ such that $v_n < \frac{3}{4}$ for $n \geq N$ - For $n \geq 5$, $u_{n+1} < \frac{3}{4} u_n$ - For $n \geq 5$, $u_n \leq \left(\frac{3}{4}\right)^{n-5} u_5$ - $S_n \leq \left(1 + \frac{3}{4} + ... + \left(\frac{3}{4}\right)^{n-5}\right) u_5 \leq 4 u_5$ - $(S_n)$ is increasing and convergent.