calculus - Find an equivalent of sequence at infinity

I want to find an equivalent at infinity to those two sequences and then deduce their possible limits:

$$u_n=\frac{(-1)^n+1}{n+\sqrt{n}},\,v_n=\frac{n^5+e^n}{2n+e^n}.$$
For the first one, I found, using $n\sim \sqrt{n}$ that
$$u_n\sim \frac{(-1)^n+1}{2n},$$
and I use then the fact that both $u_{2p}$ and $u_{2p+1}$ converge to zero to obtain that $u_n$ converges also to zero. Can we find a more elegant equivalent of $u_n$ at infinity?

For the second, we already know that
$$\lim_{n\rightarrow +\infty}n^a\,e^{-b\,n}=0, a>0,b>o$$
so the limit of the second sequence $v_n$ is $1$ if we divide both the numerator and the denominator by $e^n$. But, how can we find an equivalent $w_n$ of $v_n$ such that $w_n$ converges to the value $1$ at the infinity?

Answer

$$u_n=\frac{(-1)^n+1}{n+\sqrt{n}}\\0 \leq u_n \leq \frac{1+1}{n+\sqrt{n}} \leq \frac{2}{\sqrt{n}+\sqrt{n}}\\ 0\leq u_n \leq \frac{1}{\sqrt{n}}$$ when $n \rightarrow \infty$ $$\frac{1}{\sqrt{n}} \rightarrow 0 \\ \rightarrow u_n \rightarrow 0$$