수열의 수렴과 발산

(a) Since $-1 \leq \sin n \leq 1$ for all $n$, one potential set of upper and lower bound is $$\dfrac{-1}{n}\le \dfrac{\sin n}{n}\le \dfrac{1}{n}$$ Note $\lim\limits_{n \to \infty}\dfrac{-1}{n} = \lim\limits_{n \to \infty}\dfrac{1}{n}$, so these are valid comparison sequences for the squeeze theorem. (b) Since $-1\leq \sin n \leq 1$ and $-5 \leq -5\cos n \leq 5$ for all $n$, we see

$$\begin{aligned} 7 -1 -5&\leq 7+ \sin n - 5 \cos n \leq 7+{red}1+{blue}5\\ 1&\leq 7+ \sin n - 5 \cos n \leq13 \end{aligned}$$

This gives us the idea to try the bounds $$\dfrac{n^2}{13e^n} \le \dfrac{n^2}{e^n(7+\sin n - 5\cos n)} \le \dfrac{n^2}{e^n}$$ We check that $\lim\limits_{n \to \infty}\dfrac{n^2}{13e^n} = \lim\limits_{n \to \infty}\dfrac{n^2}{e^n}$ (they're both 0–you can verify using l'Hopital's rule), so these are indeed reasonable bounds to choose to use with the squeeze theorem. Alternatively, since $0\le \frac{n^2}{13e^n}$, we can also use $$0 \le \dfrac{n^2}{e^n(7+\sin n - 5\cos n)} \le \dfrac{n^2}{e^n}$$ (c) Since $(-n)^{-n} = \dfrac{1}{(-n)^n} = \dfrac{(-1)^n}{n^n}$, we see $$\dfrac{-1}{n^n} \le (-n)^{-n} \le \dfrac{1}{n^n}$$ Since both $\lim\limits_{n \to \infty}\dfrac{-1}{n^n}$ and $\lim\limits_{n \to \infty}\dfrac{1}{n^n}$ are 0, these are reasonable bounds to use with the squeeze theorem.