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Now of course I'm not stranger to the fact that adding finite (and in many cases - infinite) amount of positive numbers always yeilds a positive number, but in many cases, often the finite limit isn't equivalent to the strange nature of infinity. It seems that mathematics tends to prefer that if $\sum_{n=0}^\infty2^n$ were to converge, it likes

$$\sum_{n=0}^\infty2^n=-1$$

This can be seen in many "proofs" like forcing the geometric series formula:

$$\sum_{n=0}^\infty r^n=\frac{1}{1-r}\implies\sum_{n=0}^\infty2^n=\frac{1}{1-2}=-1$$

Or the digit function (which returns the digit in the $b^n$ column of $x$ in base $b$):

$$D_b^n(x)=\lfloor\frac{x}{b^n}\rfloor-b\lfloor\frac{x}{b^{n+1}}\rfloor$$

(E.g; $D_{10}^{-2}(\pi)=4$)

For $-1$: $$D_2^n(-1)=1 \forall n>0$$ $$D_2^n(-1)=0 \forall n\leq0$$

Hence it could be seen that $-1_{(10)}=\bar1.0_{(2)}$, which implies $$\sum_{n=0}^\infty2^n=-1$$

Besides conventions for convergence, what exactly is stopping this sum from being equal to $-1$? Would such an equivalence lead to a contradiction?

Computationally concerned, using $-1_{(10)}=\bar1.0_{(2)}$ as a binary expansion holds all the expected arithmetic properties like $\bar1.0_{(2)}\times\bar1.0_{(2)}=\bar01.0_{(2)}$

Graviton
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If $$\sum_{n=0}^\infty 2^n$$ converges, then it converges to $-1$

is a true statement.

So is the statement

If $$\sum_{n=0}^\infty 2^n$$ converges, then it converges to $42$.

And so is the statement

If $$\sum_{n=0}^\infty 2^n$$ converges, then the Earth is flat.

All three statements are true because $A\implies B$ is always true if $A$ is false. The simple fact is that the expression $$\sum_{n=0}^\infty 2^n$$ denotes the sum of a divergent series, and it therefore cannot be used in an equality because it is undefined.

The expression $$\sum_{n=0}^\infty a_n$$ is simply shorthand for $$\lim_{N\to\infty} \sum_{n=0}^Na_n.$$

If you plug in $a_n=2^n$ into that expression, you see that what you are really asking is why

$$\lim_{N\to\infty} \sum_{n=0}^N 2^n=-1$$ is not true. And it is not true because the right-hand side of the equality is a real number, while the left hand side is an undefined expression.

5xum
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  • I understand, so it is not equal because of the convention that an infinite summation is equal to it's limit? – Graviton Jul 22 '20 at 11:51
  • However I'm just not sure that $-1=\bar1_{(2)}$ being true actually leads to a contradiction outside of this convention. – Graviton Jul 22 '20 at 11:53
  • @Graviton That's not a convention, that's a definition. – 5xum Jul 22 '20 at 11:56
  • Indeed, though a definition that is conventionally accepted. – Graviton Jul 22 '20 at 11:57
  • @Graviton If the equality were true, then a false statement is true. That is in itself a contradiction. Further, as I said, if that statement is true, then all statements are true. Let's say $A$ is the statement $-1=\sum_{i=0}^\infty$. Now, take any statement $p$. Then, the statement $A\implies p$ is true, because $A$ is false. But also, we have that $A$ is true, and $A\implies p$ is true, so, by modus ponens, $p$ is true. QED. Even if $p$ is "the moon is made of cheese", we managed to prove $p$. – 5xum Jul 22 '20 at 11:58
  • I am quite aware of the principle of explosion, I just do not believe outside that definition for an infinite sum, $(\top\implies(-1=\bar1_{(2)}))$ is false. – Graviton Jul 22 '20 at 12:04
  • @Graviton What do you mean by "outside that definition"? Are we throwing away that definition then? If so, what do we replace it with? What does the expression $$\sum_{n=0}^\infty 2^n$$ even mean, if we just discarded the definition? It's now just a bunch of symbols with no more meaning behind it than a flat Earther congress. – 5xum Jul 22 '20 at 12:07
  • I imagine it would depend if by $\infty$ one is referring to $\omega$ or some other strictly defined infinity. The use of $\infty$ there is meaningless without specification of which infinity. – Graviton Jul 22 '20 at 12:10
  • "The use of $\infty$ there"... Where? In the summation sign? That symbol a part of a strictly defined symbol. The collection of symbols $$\sum_{n=0}^\infty a_n$$ is defined as $$\lim_{N\to\infty}\sum_{n=0}^N a_n.$$ Further, the collection of symbols $$\lim_{n\to\infty} b_n$$ is defined as the value $L\in\mathbb R$ for which the following statement is true:

    $$\forall\epsilon > 0\exists N_0\in\mathbb N: n>N_0\implies |L-b_n|<\epsilon.$$

    All symbols in the statement are again strictly defined.

     – 5xum Jul 22 '20 at 12:17
  • Let us continue this discussion in chat. – Graviton Jul 22 '20 at 12:22