Prove $\lim\limits_{n \to \infty}\frac{2^n}{n^2}+(-1)^n = \infty$

Question

$$\lim\limits_{n \to \infty}\frac{2^n}{n^2}+(-1)^n = \infty$$ I have to show this limit equals to positive infinity using these results:

• (a) Show that if $ \lim s_n = +\infty $ and $ \inf \{t_n : n \in \mathbb{N}\} > -\infty $, then $ \lim(s_n + t_n) = +\infty $.
• (b) Show that if $ \lim s_n = +\infty $ and $ \lim t_n > -\infty $, then $ \lim(s_n + t_n) = +\infty $.
• (c) Show that if $ \lim s_n = +\infty $ and if $ (t_n) $ is a bounded sequence, then $ \lim(s_n + t_n) = +\infty $.

I know $\lim\limits_{n \to \infty}\frac{2^n}{n^2} = \infty$ from another result, but I don't understand what it means to be $\gt-\infty$ for (a) and (b) and I can't use (c) because $(-1)^n$ is not a bounded sequence. So I can't make sense of $\lim\ (-1)^n$.

Answer 1

Hint: Since $(-1)^n$ is bounded, by $c)$ it is enough to show that $\frac{2^n}{n^2}$ goes to infinity as $n$ increases.

EDIT: Looks like you already know that $\frac{2^n}{n^2}$ goes to infinity as $n$ increases. To see that $(-1)^n$ is bounded notice that for any $n,$ the value of $(-1)^n$ bounces between $-1$ and $1$. Thus, its absolute value is never bigger than $1$. Thus, it is a bounded sequence.