Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can anyone explain me what the result of $$\lim_{n\rightarrow\infty} (-1)^n$$ is and the reason?

share|cite|improve this question
The limit doesn't exist. –  Qiaochu Yuan Jul 2 '11 at 21:04

5 Answers 5

up vote 12 down vote accepted

This does not exist. If you have a sequence $\{x_n\}$, then if $x_n \rightarrow l$, for any open interval $I$ with $l\in I$, $x_n\in I$ for all but finitely many $n$. Intuitively, any open interval containing the limit must "eventually absorb" the sequence.

Your sequence has no such behavior. If you take the interval $(.9, 1.1)$, we hve $x_n\not\in I$ if $n$ is odd. Likewise, taking a small interval around $-1$ results in $x_n$ failing to be in that interval if $n$ is even. There is no point you can pick to eventually absorb the sequence, and therefore there is no limit.

share|cite|improve this answer
Thanks for the precise answer - helped me out! (I will accept this as the solution when i am allowed in about 5 minutes) –  Christian Ivicevic Jul 2 '11 at 21:10

A limit of a sequence exists when there is a number $L$ such that for every $\epsilon>0$ there is some $N\in\mathbb N$ such that for every $n>N$ we have $|a_n-L|<\epsilon$.

The limit is infinite if for every $M>0$ there is some $N\in\mathbb N$ such that for all $n>N$ we have $a_n>M$.

In this case $a_n=(-1)^n$, which takes two values: $1,-1$.

We do not even need our counterexample $\epsilon$ to be small, just set $\epsilon=1$. Then for every number $L$ we have:

If $|1-L|<1$ then $|-1-L|\ge 1$, and if $|-1-L|<1$ then $|1-L|\ge 1$.

Therefore for every $N\in\mathbb N$ either $a_{N+1}$ or $a_{N+2}$ is of at least distance of $1$ from $L$, for any given $L$.

The limit cannot be infinite either, for obvious reasons.

We are therefore left only with the possibility that the limit does not exist.

share|cite|improve this answer

If a sequence converges all its subsequences converge to the same limit.

Note that $(-1)^{2n}$ is a subsequence that converges to $1$ and $(-1)^{2n + 1}$ a subsequence that converges to $-1$. Contradiction.

share|cite|improve this answer

The limit is undefined. Since (-1)^(2n+1)=-1 & (-1)^(2n+1)=1, raising to n where approaches infinity means that you can't define whether the result is +1 or -1.

share|cite|improve this answer
Welcome to math.stackexhange. Note that this question is over two years old, and the answers given already contains what you wrote. It might be better to address a newer question that doesn't already have a satisfactory answer. –  mrf Nov 16 '13 at 21:01

A humble engineer's thought: think about what $(-1)^t$ for $t\in\mathbb{R}$ would be if you let $t\rightarrow\infty$. When you substitute $(-1)=\mathrm{e}^{-\mathrm{j}\pi}$ you see that the expression is actually just a complex phasor, which keeps spinning around the origin. The limit is thus undefined. Maths experts, please don't kill me :)

share|cite|improve this answer
A cute way to link this POV with the subsequence comments in the other answer: If I were to treat $t$ as a time variable and 'sampled' the position of the phasor at a rate of $2$, then the phasor would appear to stand still since the motion is $2$-periodic (i.e. there's a stroboscopic effect). But this position is determined by when I initially start sampling, so there can't be a well-defined limit. –  Semiclassical Jul 7 at 16:19

protected by Zev Chonoles Jul 7 at 15:26

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

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