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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?

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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.

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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.

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  • $\begingroup$ I don't understand why this works. One might well say that & (-2)^n \leq 2^n& and show that $2^n$ approaches infinity, but this does not mean that $(-2)^n$ does. $\endgroup$ – Aemilius Oct 19 '17 at 6:27
  • $\begingroup$ $\lim_{n \to \infty} f(n) = -\infty$ means that given any number $M$, we can pick a number $N>0$, such that if $n>N$, then $f(n) < M$. If $1-n$ can be arbitarily negative, a quantity that is a lower bound of it can be arbitrarily negative as well. $\endgroup$ – Siong Thye Goh Oct 19 '17 at 6:30
  • $\begingroup$ But what about my example? $\endgroup$ – Aemilius Oct 19 '17 at 6:32
  • $\begingroup$ We have to bound things in the right direction, if we have $g(n)<2^n<h(n)$, if we can show that $2^n$ goes to $\infty$, we can't conclude that $g(n)$ goes to $\infty$, we can conclude that $h(n)$ being its upper bound goes to $\infty$ as well. $\endgroup$ – Siong Thye Goh Oct 19 '17 at 6:32
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Hint $$-1-n\le (-1)^n-n \le 1-n$$ Since

$$-1\le (-1)^n \le 1$$

so $$-\infty =\lim_{ n \to \infty } -1-n\le \lim_{ n \to \infty } (-1)^n-n \le \lim_{ n \to \infty } 1-n=-\infty $$

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  • $\begingroup$ Can we apply the squeeze theorem with sequences which don't have a proper limit? $\endgroup$ – Aemilius Oct 19 '17 at 7:08
  • $\begingroup$ yes as far both extrême limits are equals. but in case of infties in not necessary. ok side is usually enough $\endgroup$ – Guy Fsone Oct 19 '17 at 7:13
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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 $n>N$.

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  • $\begingroup$ But what is N(c)? $\endgroup$ – Aemilius Oct 19 '17 at 7:06
  • $\begingroup$ Definition: a sequence $(a_n)$ of real numbers converges to $ - \infty \iff $ to each $c<0$ there is an index $N=N(c)$ such that $a_n <c$ for all $n>N$. $\endgroup$ – Fred Oct 19 '17 at 7:30

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