Is $(-1)^{\infty}=0$?

Question

I was recently working across the ADT Queue when I thought that such a scenario can be defined by the Grandi series. It is given that: $$ \sum_{n=0}^k (-1)^n = \frac{1}{2}\big((-1)^k + 1\big) $$ It is seemingly visible that both are divergent series, and as $k \rightarrow \infty$ it can be said that: $$ \sum_{n=0}^\infty (-1)^n = \frac{1}{2}\big((-1)^\infty + 1\big) $$ It can be now said that: $$ 2\sum_{n=0}^\infty (-1)^n = (-1)^\infty + 1 $$ The LHS gives us the infinite series:

2-2+2-2+2...

It can now be solved on LHS, treating it as divergent series: $$ \begin{gather} S = 2-2+2-2+2...\\ 2-S = 2-(2-2+2-2+2...)\\ 2-S=S\\ S=2S\\ S=1\\ \end{gather} $$ Continuing where we left above solving the LHS, it can be said, $$ 1 = (-1)^{\infty} + 1 $$ $$ (-1)^{\infty}=0 $$ My question is:

• Is $(-1)^{\infty}$ actually determinate?
• Is the proof I wrote here correct?

Disclaimer: I am newly 16 years old elementary math student with no expertise (in formal ways) in the subject. I did not write this question simply to waste others' time. I hope the question stays here, with a criticism/appreciation of the proof I presented.

Answer 1

You have proven:

If $\displaystyle\sum_{n=1}^\infty (-1)^n$ converges, then $\displaystyle\lim_{n\to\infty}(-1)^n = 0$.

There is a quicker way to get there: The terms in any convergent series must tend to zero as a sequence.

While the conditional statement is true, its hypothesis is not satisfied: Grandi's series does not converge. So no conclusion can be drawn from the statement.

Answer 2

$(-1)^x$ is usually not defined for non integer $x$. So here stating that $(-1)^\infty$ exists is like saying that $\infty$ is an integer. This is not true, as $\infty$ is not a real or complex number. Another argument we could make is that you have "rearranged" the terms in your sum. This changes its value according to Riemann's rearrangement theorem, which states that shuffling the terms of a convergent sum could change its value.

Answer 3

No, it cannot be said.

The limit of $(-1)^n$ as $n\to \infty$ doesn't exist.

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