How to prove $x^n$ has no limit for $x\leq -1$

Question

Writing out the first few terms for $x=-1$:$$(-1)^1,(-1)^2,(-1)^3,...,(-1)^n$$ we can clearly see that the terms alternate between positive and negative and that since $-1\nleq x\ngeq1$it is clear that the sequence does not converge. My question is that how do I sum up my ideas into a rigorous proof?

Answer 1

HINT

Let consider what appens for $n$ even and for $n$ odd. Are the limits in these two cases equals?

Recall that for any subsequence $a_{i_n}$ of $a_n$

$$a_n\to L \implies a_{i_n}\to L$$

Answer 2

Using the definition of convergence is, in my opinion, the best way to make this rigorous. If a sequence $a_n$ converges to some limit $L$, then by definition of the word "converges", it means that for any $\varepsilon>0$ there is an $N\in \Bbb N$ such that for any $n>N$ we have $$ |a_n-L|<\varepsilon $$ Thus in order to disprove convergence, you have to prove the negation. It is a small exercise in logic to negate the above correctly, but here is, by definition, what it means for a sequence $a_n$ to not converge to some $L$:

There is an $\varepsilon > 0$ such that for any $N\in \Bbb N$, there is an $n>N$ where $$ |a_n - L|\geq \varepsilon $$

In order to show that it doesn't converge at all, you must show that this is true for all $L$. In this case, picking something like $\varepsilon = \frac14$ works for all $L$.

Answer 3

For $(-1)^n$: Suppose it is converges to $l$ say. Then $$\vert (-1)^n -l \vert < \frac{1}{2}$$ for all $n \geq N$.

For $n$ even, this says $$\vert 1 -l \vert < \frac{1}{2}$$ for all $n \geq N$

For $n$ odd, this says $$\vert -1 -l \vert < \frac{1}{2}$$ for all $n \geq N$

Now $$2=\vert 2 \vert =\vert 1+1\vert =\vert (1+l)+(1-l) \vert<1$$ which is absurd!