<s>[INST] How to prove $\lim_{n \to \infty} \frac{n}{1+\sqrt{n}}cos(\frac{\pi n}{n+1})= -\infty$? I tried to prove by using definition for divergence:

We say that a sequence $a_n$ diverges to $-\infty$ if for every $M>0$, there is an integer $n_0$ such that $a_n < -M$ whenever $n>n_0$

So I start:
$$\frac{n}{1+\sqrt{n}}cos(\frac{\pi n}{n+1}) < -M$$
Because $|cos x|\leq 1:$
$$-\frac{n}{2} \leq -\frac{n}{1+\sqrt{n}} \leq \frac{n}{1+\sqrt{n}}cos(\frac{\pi n}{n+1}) < -M$$
$$-\frac{n}{2} < -M$$
$$n>2M$$
Therefore $n_0$ exists and we can say (for example): $n_0=2M+1$.
Is this correct? Is there some other way we could prove this?
 [/INST] Hi I have like no idea about this being proof enough for you but we usually do this:
$
 \lim_{n \to \infty} \frac{n}{1+\sqrt{n}}\cos\left(\frac{\pi n}{n+1}\right)= -\infty
$
$
1. \frac{\pi n}{n+1} = \frac{\pi }{1+\frac{1}{n} } 
$ // divide by n
for $n\to
 \infty ,  \frac{1}{n} = 0 
$
$
\frac{\pi }{1+\frac{1}{n} } = \frac{\pi }{1+0}
$
so $\cos(\pi) = -1$  // so to get to $-\infty$  we only need to show  now that:
$ 
2. \frac{n}{1+\sqrt{n}} \to \infty  
$ // divide by n again
$ 
\frac{n}{1+\sqrt{n}} = \frac{n}{n\cdot\left( \frac{1}{n} + \frac{\sqrt{n}}{n}\right) } 
$
for
$n->  \infty, \frac{1}{n} = 0 
$
and
$  \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}}=0 
$
so we have
$
 \frac{1}{1\cdot( 0 + 0) }  = \frac{1}{0}
$ that goes to infinity
so we end with
$
 \infty \cdot -1  = -\infty$
Take care! I did not write the lim in every row because of fatal lazyness
</s>