Limes $1- \frac{(-1)^n}{\sqrt{n}}$ and $\frac{2^n + (-3)^n}{(-2)^n + 3^n}$ and $n - 3 \lfloor \frac{n}{3} \rfloor$

Question

I need to find the limes superior, limes inferior and limes (if they exist) for

$$\text{1. } a_n := 1- \frac{(-1)^n}{\sqrt{n}}, n \in \mathbb{N}$$

$$\text{2. } b_n := \frac{2^n + (-3)^n}{(-2)^n + 3^n}, n \in \mathbb{N}$$

$$\text{3. } c_n := n - 3 \lfloor \frac{n}{3} \rfloor$$

Regarding $1.$ I have that

$$ 1- \left|\frac{(-1)^n}{\sqrt{n}}-\frac{(-1)^m}{\sqrt{m}}\right|\leq 1-\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{m}}\rightarrow 1 \quad\text{as}\quad m,n\rightarrow\infty $$ It follows that $$ \lim_{m,n\rightarrow\infty}1-\left|\frac{(-1)^n}{\sqrt{n}}-\frac{(-1)^m}{\sqrt{m}}\right|=1. $$

So the $\lim \sup_{n \to \infty} a_n = 1 \text{ and } \lim \inf_{n \to \infty} a_n = 1$

Regarding $2.$ I have that

$$\lim_{n \to \infty} \frac{(-3)^n}{3^n} \frac{1-(2/3)^n}{1+(2/3)^n} = 1$$ $\lim \sup_{n \to \infty} b_n = 1 \text{ and } \lim \inf_{n \to \infty} b_n = 1$

Regarding $3.$ I have that

$$\{n\}=n-\lfloor n\rfloor\qquad\qquad 0\leq \{n\}<1$$

We get \begin{align*} \lim_{n\to\infty}\left(n-3\left\lfloor\frac n3\right\rfloor\right) &=\lim_{n\to\infty}\left(n-3\left(\frac{n}{3}-\left\{\frac{n}{3}\right\}\right)\right)\\ &=\lim_{n\to\infty}\left(3\cdot\left\{\frac{n}{3}\right\}\right)\\ &=3\cdot\lim_{n\to\infty}\left\{\frac{n}{3}\right\}\quad\not\exists\tag{1}\\ \end{align*}

From $(1)$ we can see that the values oscillate in $[0,3)$ when $n$ increases, so that the limit does not exist. Since the limits of the subsequences

\begin{align*} 3\cdot\lim_{{n\to\infty}}\left\{\frac{n}{3}\right\}=0 \qquad\qquad\text{and}\qquad\qquad 3\cdot\lim_{{n\to\infty}}\left\{\frac{n}{3}\right\}=\frac{3}{2}\\ \end{align*} are different, the limit $(1)$ does not exist. So there are no $\lim \inf_{n \to \infty} c_n $ and $\lim \sup_{n \to \infty} c_n$

I think that I have some mistakes and that what I've written is false. Can someone verify or tell me what the right solution is?

Answer 1

The first one is ok.

The second one is incorrect since $\lim \dfrac{(-3)^n}{3^n}$ doesn't exist.

Proof 2: $$b_n=\dfrac{2^n+(-3)^n}{(-2)^n+3^n}$$ For odd $n$, $b_n=\dfrac{2^n-3^n}{-2^n+3^n}=-1$.

For even $n$, $b_n=\dfrac{2^n+3^n}{2^n+3^n}=1$

Therefore, $\lim\sup b_n=1, \lim\inf b_n=-1$

The third one has a problem. The sequence oscillate in $\{0,0.\dot3,0,\dot6\}$ since $n\in\mathbb{N}$.

So, $\lim\sup c_n=3\times\dfrac{2}{3}=2,\lim\inf c_n=0$