1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What do you mean for "divergent limit"? $\endgroup$ – Alessio Ranallo Nov 26 '15 at 15:36
  • 1
    $\begingroup$ Does this help? math.stackexchange.com/questions/233215/… $\endgroup$ – Brenton Nov 26 '15 at 15:45
  • 2
    $\begingroup$ 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. $\endgroup$ – Guest Nov 26 '15 at 15:46
  • $\begingroup$ Hint: let $x>0$, so $-x<0$ then you can write $-x=(-1)(x)$ $\endgroup$ – Alessio Ranallo Nov 26 '15 at 15:48
1
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.