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

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

5 Answers 5

up vote 11 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.

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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

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.

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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.

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The limit does not exist.

$(-1)^n$ is $-1$ when $n$ is odd and $1$ when $n$ is even. As $\infty$ is both odd and even, the "limit" is both $-1$ and $1$. Therefore, the limit does not exist.

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I think this explanation is nonsense. –  GEdgar Jul 2 '11 at 22:03
At least the double-parity of infinity has been invoked! –  The Chaz 2.0 Jul 3 '11 at 0:19
Well apparently infinity is even... –  BlueRaja - Danny Pflughoeft Jul 3 '11 at 4:08

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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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

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