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

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    $\begingroup$ The limit doesn't exist. $\endgroup$ – Qiaochu Yuan Jul 2 '11 at 21:04

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.

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

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.


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.


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 :)

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    $\begingroup$ 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. $\endgroup$ – Semiclassical Jul 7 '15 at 16:19

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.

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    $\begingroup$ 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. $\endgroup$ – mrf Nov 16 '13 at 21:01

protected by Zev Chonoles Jul 7 '15 at 15:26

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