How to prove $\lim_{n \to \infty} \frac{n}{1+\sqrt{n}}cos(\frac{\pi n}{n+1})= -\infty$?

Question

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?

Answer 1

hint

Sorry, your first inequality is not correct.

The only way is to write

$$\cos(\frac{\pi n}{n+1})=\cos(\pi-\frac{\pi}{n+1})$$ $$=-\cos(\frac{\pi}{n+1})$$

And

$$\frac{n}{1+\sqrt{n}}\ge \frac{n}{2\sqrt{n}}$$

But, for $ n $ great enough $(n\ge N_1)$, we have $$-\cos(\frac{\pi}{n+1})<-\frac 12$$ since it goes to $ -1$. thus

$$n>N_1\;\implies \; a_n<-\frac{\sqrt{n}}{4}$$ Given $ M>0 $

You just need to find $ N_2 $ such that $$n> N_2 \;\implies \; \frac{\sqrt{n}}{4}>M$$

You can take $ N_2=16M^2$.

Finally, if you take $ N=\max(N_1,N_2) $, then

$$n>N\;\;\implies \;\;a_n<-M$$

Answer 2

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