Find $\sup\{u_n\}$ and $\inf\{u_n\}$, where
1) $u_n=(-1)^n+\cos(\frac{n\pi}{4})$
2) $u_n=\frac{(-1)^n}{n}+\sin(\frac{n\pi}{2})$
My answer:
Let $x_n=(-1)^n$ and $y_n=\cos(\frac{n\pi}{4})$ $\forall n\in\mathbb{N}$.
Then, $\{x_n\}=\{-1,1,-1,1,-1,1,-1...\}$
and $\{y_n\}=\{\frac{1}{\sqrt{2}},0,\frac{-1}{\sqrt{2}},-1,\frac{-1}{\sqrt{2}},0,\frac{1}{\sqrt{2}},...\}$.
$\therefore$ range of $\{u_n\}=\{(-1+\frac{1}{\sqrt{2}}),1,(-1-\frac{1}{\sqrt{2}}),0\}$
$\Rightarrow\sup\{u_n\}=1$ and $\inf\{u_n\}=(-1-\frac{1}{\sqrt{2}})$Similarly, we find range of $\{u_n\}=\{(-1+1),(\frac{1}{2}+0),(\frac{-1}{3}-1),(\frac{1}{4}+0),(\frac{-1}{5}+1)...\}$.
Define sequences $\{a_n\},\{b_n\},\{c_n\}\backepsilon$
$\{a_n\}=\{(-1+1),(-\frac{1}{5}+1),(-\frac{1}{9}+1),...\}\in[0,1)\forall n\in\mathbb{N}$ and $\sup\{a_n\}=1$
$\{b_n\}=\{\frac{1}{2},\frac{1}{4},\frac{1}{6},...\}\in(0,\frac{1}{2}]\forall n\in\mathbb{N}$
$\{a_n\}=\{(-1-\frac{1}{3}),(-1-\frac{1}{7}),...\}\in[-1-\frac{1}{3},-1)\forall n\in\mathbb{N}$ and $\inf\{c_n\}=-1-\frac{1}{3}$
$\therefore\sup\{u_n\}=1$ and $\inf\{u_n\}=-1-\frac{1}{3}$
Questions: Is there an alternative method which doesn't rely totally on finding the range to get $\sup,\inf$ for such sequences?
Also, what can be said about the convergence of these sequences?
For each $n \in \mathbb{N}$, as $$ -1 \leq \cos \frac{n \pi}{4} \leq 1, $$ and as $$ -1 \leq (-1)^n \leq 1, $$ so we must have $$ -2 = -1 + (-1) \leq (-1)^n + \cos \frac{n \pi}{4} \leq 1 + 1 = 2. $$
Thus the sequence $\left( u_n \right)_{n \in \mathbb{N}}$ is indeed a bounded sequence, and we have the inequalities $$ -2 \leq \inf \left\{ \ u_n \, \colon \, n \in \mathbb{N} \ \right\} \leq \sup \left\{ \ u_n \, \colon \, n \in \mathbb{N} \ \right\} \leq 2. $$
However, we note that the terms of this sequence corresponding to $n = 1, 2, 3, \ldots, 8$, respectively, run as follows: $$ \begin{matrix} n & = & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & \ldots \\ u_n & = & -1+\frac{1}{\sqrt{2}} & 1 & -1- \frac{1}{\sqrt{2}} & 0 & -1 - \frac{1}{\sqrt{2}} & 1 & -1+\frac{1}{\sqrt{2}} & 2 & \cdots \end{matrix} $$
And, for any $n \in \mathbb{N}$, we find that $$ u_{n+8} = (-1)^{n+8} + \cos \frac{(n+8) \pi }{4} = (-1)^n(-1)^8 + \cos \left( \frac{n \pi }{4} + 2\pi \right) = (-1)^n + \cos \frac{ n \pi}{4} = u_n. $$
Therefore the set $$ \left\{ \ u_n \, \colon \, n \in \mathbb{N} \ \right\} = \left\{ \ -1 - \frac{1}{\sqrt{2}}, -1+\frac{1}{\sqrt{2}}, 0, 1, 2 \ \right\}. $$
Hence $$ \inf \left\{ \ u_n \, \colon \, n \in \mathbb{N} \ \right\} = -1 - \frac{1}{\sqrt{2}} $$ and $$ \sup \left\{ \ u_n \, \colon \, n \in \mathbb{N} \ \right\} = 2. $$
Hope this helps.
Can you now handle the other sequence similarly?