Thinking about the sum of all integers from $-n$ to $n$, which is $$ - \frac {n(n+1) } {2} + \frac {n(n+1) } {2} = 0,$$ got me into confusion. Why, extending this to the whole integer set $\mathbb Z$, do we get the indeterminate form $\infty - \infty $? Why is that? Can we prove somehow this indeterminate form statement? To me it feels unnatural that summing the all integer set becomes a divergent operation, instead of simply zero.
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3Insight, intuition, rigor: develop all three – A rural reader Aug 04 '24 at 20:02
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1arithmetic operations are not even defined for $\infty$. Consider $x^2 - x$, which goes to infinity as $x$ goes to infinity. – paulina Aug 04 '24 at 20:07
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2Because the value of the sum depends on the order of summation. – copper.hat Aug 04 '24 at 20:08
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In most applications the answer would be $0$, as this is the (Cauchy) Principle Value, precisely by the argument you outlined. – Brevan Ellefsen Aug 04 '24 at 20:26
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1You have preferred $0$ as the middle. But if you paired the inters as $$(0+1)+((-1)+2)+\cdots+((-n)+(n+1))+\cdots$$ then the sum becomes $1+1+1+\cdots.$ – Thomas Andrews Aug 04 '24 at 20:40
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A sum like $\sum\limits_{n = -\infty}^\infty n$ is defined as $\lim\limits_{\substack{a \to \infty \\ b \to \infty}} \sum\limits_{n = -a}^b n$ rather than $\lim\limits_{{c \to \infty}} \sum\limits_{n = -c}^c n$. Yes, the latter would be $0$ but it's not a robust definition.
You're imagining putting $0$ in the middle and then adding pairs on either side like
$$ 0 + (-1 + 1) + (-2 + 2) + (-3 + 3) + \cdots = 0. $$
But you could just as well start with $1$ in the middle instead of $0$ and get
$$ 1 + (0 + 2) + (-1 + 3) + (-2 + 4) + (-3 + 5) + \cdots = \infty. $$
We want a definition of limit which doesn't depend on which digit is the middle digit.
Sera Gunn
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You are right,
$$0+\sum_{n=1}^\infty\left(-n+n\right)=0+\sum_{n=1}^\infty0=0$$
and this is equivalent to writing
$$\lim_{n\to\infty}\left(-\frac{n(n+1)}2+\frac{n(n+1)}2\right)=\lim_{n\to\infty}0=0.$$
Problem is that this does not describe "the sum of all integers", because it could be done differently, as said by others.