$$\sum_{n=1}^\infty (-1)^n$$ Is this mathematical expression zero or undefined? I think it looks like zero but i can't explain the reason mathematically. In addition, $\infty - \infty$ is undefined afaik?
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The series is irregular (neither converges nor diverges) and you could say the sum is undefined. – Vincenzo Oliva Sep 22 '16 at 18:01
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I assume that was a typo. – Andres Mejia Sep 22 '16 at 18:04
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In addition to the (correct!) answers that this limit is undefined, your intuition that it "looks like zero" is not completely wrong. You may find the following article interesting: https://en.wikipedia.org/wiki/Banach_limit – sTertooy Sep 22 '16 at 18:12
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despite this sum isn't finite in a conentional sense, it is $1/2$ in any well defined regularisation scheme. For example using Abel or Cesaro summation... – tired Sep 22 '16 at 18:46
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@tired You probably mean $-1/2$. – Did Sep 22 '16 at 19:20
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@Did sure, thanks for spotting – tired Sep 22 '16 at 19:40
2 Answers
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It is undefined. The sequence of partial sums is $-1, 0, -1, 0...$ which, while bounded, does not converge.
basket
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As a general rule, if $\sum a_n$ converges, it must be the case that $a_n \to 0$. In the case of this series, $(-1)^n=-1,0,-1,0$ doesn't converge to anything,so it certainly does not tend to $0$, and so you can deduce that the series is divergent.
For your last comment, referring to $\infty-\infty$ suggests to me that you are doing the following: $$\sum_{n=1}^{\infty} (-1)^n=\sum_{n=1}^{\infty} (-1)^{2n}-\sum_{n=1}^{\infty} (-1)^{2n+1} ``=" \infty-\infty.$$
However, you can only re-order infinitely many terms in a series (assuming you want a sensible result) if the series is absolutely convergent, which $\sum_{n=1}^{\infty}(-1)^n$ is not, since $$\sum_{n=1}^{\infty} |-1^n|=\sum_{n=1}^{n} 1^n=1+1+1+...$$
Andres Mejia
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