Calculate $\lim_{n \to \infty} n(1+\sin(n))$ I think it doesn't exists but I can't find out how to prove it; note that $n$ is integer.
I know that there are infinite many integers $a_n$ s.t. $\sin(a_n)>0$ and then limsup is $+ \infty$.
I want to prove that liminf $\ne +\infty$ but I don't know how.
 A: The limit doesn't exist. We are going to prove it the hard way.


*

*$\limsup\limits_{n\to\infty}n(\sin n +1 ) = +\infty$
If one make a plot of $\max( \sin(x),\sin(x+3) )$ over $[0,2\pi]$, one will notice it is bounded from below by $-2\cos(1.5) \approx -0.0707372$.
For any $n$, at least one of $\sin n + 1$ or $\sin(n+3) + 1$ will be greater than $0.9$. As a result
$$\limsup_{n\to\infty} n(\sin n+1) \ge \limsup_{n\to\infty} 0.9n = \infty$$

*$\liminf_\limits{n\to\infty} n(\sin n+1) = 0$.
We will use a theorem by Minkowski,

Given any $\theta \notin \mathbb{Q}$ and $\alpha \notin \mathbb{Z}$ such that
  $x - \theta y - \alpha = 0$ has no integer solutions.
  For any $\epsilon > 0$, there exists infinitely many pairs of integers $p,q$ such that 
  $$|q(p - \theta q - \alpha)| < \frac14\quad\text{ and }\quad |p - \theta q - \alpha| < \epsilon$$

Apply this to $\theta = \frac{1}{2\pi}$ and $\alpha = -\frac34$, there are infinitely many pairs of non-zero $p,q$ such that
$$\left|q - 2\pi\left(p+\frac34\right)\right| < \frac{\pi}{2|q|}$$
If $q$ is positive, let $n = q, m = p$. If $q$ is negative, let $n = -3q$ and $m = -3(p+1)$.
Above inequality reduces to
$$\left|n - 2\pi\left(m + \frac34\right)\right| < \frac{\pi}{2n} \times
\begin{cases}1, & q > 0\\ 9, & q < 0\end{cases}$$
Together with another easy to verify inequality, 
$$0 \le \sin\left(\frac{3\pi}{2}+\theta\right) + 1 \le \frac{\theta^2}{2}$$
we obtain infinitely many positive $n$ such that
$$0 \le n(\sin(n) + 1) \le \frac{81\pi^2}{8n}$$
This implies $$\liminf_{n\to\infty} n(\sin n+1) \le \liminf_{n\to\infty} \frac{81\pi^2}{8n} = 0$$
Since that limit inferior has to be non-negative, it has to be zero.
Combine $1)$ and $2)$, we can conclude $\;\lim_{n\to\infty}n(\sin n + 1)\;$ doesn't exist.
Notes
About this particular theorem (Minkowski has tons of theorems), 
Minkowski first proved it around $1901$ using some geometrical methods he invented. Nowadays, those methods are known as Geometry of numbers. Look at any
book under this subject, you should find a proof of that. 
I learned this stuff from a book The Geometry of Numbers by Olds, Lax and Davidoff's. It is easy to read and has a proof at Chapter 10.
Another proof can be found at Chapter 2 of Ivan Niven's classic Diophantine Approximations.
A: consider $$u_n=v_n.w_n $$ with $\lim v_n=+\infty $ and $\lim w_n $ doesn't exist, so to compute $\lim u_n $, we get an ideterminate form.
$$\{\sin (n),n\in\mathbb N\} $$ is dense in $[-1,1] $ .
