Prove that $(-1)^n - n$ is divergent to $-\infty$ Prove that $(-1)^n - n$ is divergent to $-\infty$
I tried to solve this problem. My idea was to show that the sequence indeed does approach $-\infty$ for both odd and even arguments. Now, the problem is, is it enough to say that a sequence is divergent if these two subsequences (which constitute the whole sequence after all) are divergent? How should I put the proof together? 
 A: Guide:
$$(-1)^n-n \leq 1-n$$
Just show that $1-n \to -\infty $ as $n\to \infty$.
Remark:
We can say that the sequence is divergent, but we can't conclude it goes to $-\infty$ just by looking at two subsequence.
A: Hint
$$-1-n\le (-1)^n-n \le 1-n$$
Since 
$$-1\le (-1)^n \le 1$$
so
$$-\infty  =\lim_{  n    \to \infty  }  -1-n\le   \lim_{  n    \to \infty  } (-1)^n-n \le \lim_{  n    \to \infty  }   1-n=-\infty  $$
A: Let $c<0$. Then there is $N=N(c)$ such that $-n <c-1$ for all $n>N$.
Then we get: $(-1)^n-n<c$ for all $n>N$.
A: Your approach can be completed as follows. 
You must show
$$ \forall M\in \mathbb{R}\:\ \exists N \in \mathbb{N} \: \forall n \ge N : x_n \leq M. \tag{1} $$
Your investigation of the even subsequence has yielded
$$ \forall M\in \mathbb{R}\:\ \exists K_1 \in \mathbb{N} \: \forall k \ge K_1 : x_{2k} \leq M. \tag{2} $$
Your investigation of the odd subsequence has yielded
$$ \forall M \in \mathbb{R}\:\ \exists K_2 \in \mathbb{N} \: \forall k \ge K_2 : x_{2k+1} \leq M \tag{3} $$
We now establish $(1)$ as follows. Let $M \in \mathbb R$ be given. We claim that $$N = \max\{2K_1, 2K_2 + 1\}$$ will suffice. Therefore, let $n \ge N$ be given. Then $n$ is either even or odd. If $n=2k$, then $k \ge K_1$ and $x_n \leq M$ by $(2)$. If $n=2k+1$, then $k \ge K_2$ and $x_n \leq M$ by $(3)$.
This complete your approach with all the rigour that is required.

Remark: Other solutions have been suggested which do not consider subsequences. They are safer to use in general. Investigating subsequences is a dangerous general practice as we risk forgetting elements.
