# Prove that $x^n$ diverges when $x<-1$

I'm trying to solve the following exercise from Tao's Analysis 1 textbook:

"Let $x$ be a real number. Then $\lim_{n\to\infty} x^n$ exists and is equal to zero when $|x|<1$, exists and is equal to $1$ when $x=1$, and diverges when $x=-1$ or when $|x|>1$."

DEF.(Limits of sequences) If a sequence $(a_n)_{n=m}^\infty$ converges to some real number $L$ we say that $(a_n)_{n=m}^\infty$ is convergent and that its limit is $L$; we write $\lim_{n\to\infty}a_n=L$. If a sequence $(a_n)_{n=m}^\infty$ is not converging to any real number $L$, we say that the sequence $(a_n)_{n=m}^\infty$ is divergent and we leave $\lim_{n\to\infty} a_n$ undefined.

I've been able to prove the statement for $|x|<1$, $x=1$, $x=-1$, $x>1$, but I haven't been able to do the same for the case $x<-1$, so I would appreciate any hint about how to do this last case.

(NOTE: the concept of subsequence is introduced later in the textbook, so it cannot be used to solve this exercise)

Best regards,

lorenzo.

• What do you mean for "divergent limit"? – Alessio Ranallo Nov 26 '15 at 15:36
• Does this help? math.stackexchange.com/questions/233215/… – Brenton Nov 26 '15 at 15:45
• Maybe: If $x<-1$, let $y:=-x$. Then $y>1$ so that $y^2>1$ and the subsequence $x^{2n}=\left(y^2\right)^n$ diverges as you have already shown. Hence the whole sequence diverges, because we know that in general a sequence converges iff each of its subsequences converges, or put differently, a sequence diverges iff one of its subsequences diverges. – Guest Nov 26 '15 at 15:46
• Hint: let $x>0$, so $-x<0$ then you can write $-x=(-1)(x)$ – Alessio Ranallo Nov 26 '15 at 15:48

## 1 Answer

Let's assume for contradiction that $\lim_{n \to \infty}(-x)^n$ converges to $l$ for $x>1$, then $\lim_{n \to \infty}|(-x)^n| = |l|$ (I hope you can use this). Finally, $|(-x)^n| = |x^n|$ $\implies$ $\lim_{n \to \infty}x^n = |l|$, which is a contradiction, because $\lim_{n \to \infty}x^n$ does not converge when $x>1$ (as you has proved before).