sequence and comparison test is the sequence 

$${ a }_{ n }=\frac { { \left( -1 \right)  }^{ n+1 }n }{ n+\sqrt { n }  } $$ convergent or divergent?

I used comparison test, and I compared it with $\frac{n}{n+\sqrt{n}}$. As $\lim _{ n\rightarrow \infty  }{ \frac { n }{ n+\sqrt { n }  } =1 } $. So ${ a }_{ n }=\frac { { \left( -1 \right)  }^{ n+1 }n }{ n+\sqrt { n }  } $ is convergent. Is this correct?
 A: This is not correct. One may observe that
$$
\left.\frac{(-1)^{n+1}n}{n+\sqrt{n}}\right|_{n=2p}=\frac{-2p}{2p+\sqrt{2p}} \to -1 \quad \text{as}\quad p \to \infty
$$and $$
\left.\frac{(-1)^{n+1}n}{n+\sqrt{n}}\right|_{n=2p+1}=\frac{2p+1}{2p+1+\sqrt{2p+1}} \to 1 \quad \text{as}\quad p \to \infty
$$
The given sequence is not convergent.
A: No the sequence does not converge. Consider, along even numbers $n = 2k$, we have 
$$\frac{(-1)^{2k+1}(2k)}{2k + \sqrt{2k}} = - \frac{1}{1 + \frac{1}{\sqrt{2k}}} \to -1 \,\,\, \text{ as } k \to \infty$$ where the first equality comes from multiplying the top and bottom by $2k$. Whereas, along odd numbers $n = 2m + 1$, we have $$\frac{(-1)^{2m+2}(2m+1)}{2m+1 + \sqrt{2m+1}} = \frac{1}{1 + \frac{1}{\sqrt{2m+1}}} \to 1 \,\,\, \text{ as } m \to \infty$$ where again, the first inequality comes from multiplying the top and bottom by $2m+1$. Since there are two subsequences which converge to different limits, the sequence does not converge. 
A: The absolute values of the terms are $ \dfrac n {n+\sqrt n} $, and those approach $1$ as $n\to\infty$.  The factor $(-1))^n$ alternates between $1$ and $-1$, so you have the even-numbered terms approaching $1$ and the odd-numbered terms approaching $-1$.  Consequently the whole sequence does not approach a particular number.
