<s>[INST] Bounded sequences. We have the following sequences:
$a_n=n+sin(n+3^n)$
$b_n=\frac{{(-1)^n}(n+1)}{3\sqrt{n^2+7}}$
Show that $a_n$ is non bounded and $b_n$ is bounded. Also show that the subsequences $b_{2n}$ and $b_{2n-1}$ are convergent.
My work: $a_n<n+1$ and as $n\rightarrow∞$ $a_n<$∞ , so $a_n$ is non bounded
For $b_n$ , $|\frac{{(-1)^n}(n+1)}{3\sqrt{n^2+7}}|=
\frac{n+1}{3\sqrt{n^2+7}}\rightarrow\frac{1}{3}$ as $n\rightarrow∞$
so we get $0<|b_n|<\frac{1}{3}$. 
I don't know how to show that the subsequences are convergent. Maybe Weierstrass but i'm not sure.
 [/INST] Note that
$$a_n=n+\sin(n+3^n)>n-100 \to \infty$$
thus $a_n$ is not bounded and
$$-\frac{(n+1)}{3\sqrt{n^2+7}}\le b_n=\frac{{(-1)^n}(n+1)}{3\sqrt{n^2+7}}\le \frac{(n+1)}{3\sqrt{n^2+7}}$$
with $\frac{(n+1)}{3\sqrt{n^2+7}}$ bounded since $\frac{(x+1)}{3\sqrt{x^2+7}}$ is continuous and
$$\frac{(n+1)}{3\sqrt{n^2+7}}\to\frac13$$
thus for Weierstrass EVT $b_n$ is bounded.
Finally note that
$$b_{2n}=\frac{{(-1)^{2n}}(2n+1)}{3\sqrt{4n^2+7}}=\frac{(2n+1)}{3\sqrt{4n^2+7}}\to \frac13$$
$$b_{2n-1}=\frac{{(-1)^{2n-1}}(2n)}{3\sqrt{4n^2-4n+8}}=-\frac{2n}{3\sqrt{4n^2-4n+8}}\to -\frac13$$
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