Negative 1 to the power of Infinity

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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 :)

Answer 5

\begin{array}{l} ( -1)^{\infty } \ A\ very\ interesting\ concept.\\ \\ If\ you\ look\ at\ it\ from\ an\ Euler's\ formula\ point\ of\ view\ you\ get\\ \\ ( -1)^{x} =e^{i\pi x} \Longrightarrow ( -1)^{\infty } =e^{i\pi \cdot \infty } =cos( \pi \cdot \infty ) +isin( \pi \cdot \infty )\\ sin( x\pi ) =0,\ x\in \mathbb{Z}\\ Okay,\ maybe\ infinity\ is\ not\ an\ integer,\ but\ let's\ go\ with\ it...\ \\ e^{i\pi \cdot \infty } =cos( \pi \cdot \infty ) +isin( \pi \cdot \infty ) =cos( \pi \cdot \infty )\\ cos( x\pi ) =-1,\ when\ "x"\ is\ odd\\ cos( x\pi ) =1,\ when\ "x"\ is\ even\\ \\ This\ infinity\ ( \infty ) \ is\ not\ an\ ordered\ infinity,\ it\ is\ a\ random\ infinity\ so\ there\ is\ no\ way\ to\\ determine\ wether\ it\ is\ odd\ or\ even.\ So\ this\ is\ indeterminent.\\ \\ Next\ way:\\ \\ If\ you\ think\ about\ it,\ the\ Real\ Part\ of\ the\ average\ of\ f( x) =( -1)^{x} \ from\ 0\leqslant x\leqslant \infty \\ ( from\ 0\ to\ \infty ) \ should\ be\ 0\ ( zero) ,\ because\ It's\ right\ in\ between\ -1\ and\ 1.\ Well,\ obviously\\ this\ should\ be\ true,\ but\ let's\ look\ at\ this\ through\ an\ integral.\\ \\ average\ value\ of\ f( x) \ from\ point\ "a"\ to\ point\ "b"=\frac{1}{b-a} \cdot \int\limits ^{b}_{a} f( x) dx\Longrightarrow \\ We\ want\ the\ real\ part,\ because\ the\ function\ ( -1)^{x} \ when\ x\ is\ not\ an\ integer\ value\\ will\ give\ you\ complex\ numbers-Euler's\ formula\ can\ do\ that.\\ \Re ( f( x)) =Re( f( x))\\ \Re \left(\frac{1}{\infty -0}\int\limits ^{\infty }_{0}( -1)^{x} dx\right) =\Re \left(\frac{1}{\infty }\int\limits ^{\infty }_{0} e^{i\pi x} dx\right) =\frac{1}{\infty } \Re \left(\lim ^{\infty }_{0}\left[\frac{e^{i\pi x}}{2i\pi }\right]\right) =\frac{1}{\infty } \Re \left(\lim ^{\infty }_{0}\left[\frac{( -1)^{x}}{2i\pi }\right]\right) =\\ \frac{\mathbb{1}}{\infty } \Re \left(\frac{( -1)^{\infty }}{2i\pi } -\frac{1}{2i\pi }\right) =\frac{1}{\infty } \Re ( 0) =0,\ but\ this\ only\ works\ if\ ( -1)^{\infty } \ has\ a\ non\ infinite\ value.\\ \overset{\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }}{This\ doesn't\ really\ prove\ anything\ \ guess}\\ \\ \\ Now\ let's\ look\ at\ it\ another\ way:\\ \\ ( -1)^{\infty } =-1\times -1\times -1\times -1\times -1...\\ But\ you\ can\ also\ think\ of\ it\ as\\ ( -1)^{\infty } =1-2+2-2+2-2+2-2+2-2+2...\\ because\ this\ alternates\ in\ the\ same\ way.\ But\ since\ you\ can\ do\ this,\ you\ can\ do\ something\\ special!\\ \\ ( -1)^{\infty } =1-2+2-2+2-2+2-2+2-2+2...=\\ 1-2( 1-1+1-1+1-1+1-1+1...)\\ \\ This\ piece\ here\ \\ 1-1+1-1+1-1+1-1+1...\\ is\ called\ Grandi's\ series.\ It\ can\ also\ be\ expressed\ as\ an\ infinite\ series:\\ \sum ^{\infty }_{n=0}( -1)^{n}\\ This\ series\ is\ divergent,\ but\ not\ in\ the\ sense\ that\ it\ goes\ to\ infinity,\ but\ that\ it\ doesn't\ have\ \\ an\ answer.\ But\ the\ best\ answer\ people\ say\ is\ \frac{1}{2} ,\ for\ many\ good\ reasons.\\ Now,\ assuming\ that\ Grandi's\ series\ is\ equal\ to\ \frac{1}{2} ,\ we\ get:\\ \\ ( -1)^{\infty } =1-2( 1-1+1-1+1-1+1-1+1...) =1-1=0\\ This\ answer\ makes\ a\ lot\ of\ sense\ because\\ -1\cdot ( -1)^{\infty } =( -1)^{\infty } ,\ the\ only\ value\ that\ ( -1)^{\infty } \ can\ have\ now\ is\ zero:\\ k=( -1)^{\infty } ,\ k=-1\cdot ( -1)^{\infty } =-k\Longrightarrow k=-k\Longrightarrow k=0\\ \\ But,\ because\ technically\ Grandi's\ series\ doesn't\ have\ a\\ value,\ ( -1)^{\infty } \ still\ equals\ indeterminent.\\ I'm\ not\ sure\ if\ this\ helped\ at\ all,\ but\ it\ was\ fun\ for\ me\ to\ write.\ Thanks. \end{array}

I left a lot of stuff out because all the other awesome math lovers gave great details on them. I hope myy stuff helps!

Answer 6

The limit is nonexistent. Since $(-1)^{2n+1}=-1$ & $(-1)^{2n}=1$, raising to $n$ where approaches infinity means that you can't define whether the result is $+1 \text{ or} -1,$ or anything else, for thar matter.