Re: ** says: Definition: sum{i in N} i = 0

On 21 Jun., 19:59, Virgil <vir...@xxxxxxxxxxx> wrote:

So it is. And I gave three different proofs of this claim, the first
one using reciprocals, the second one being stated above and the last
one simply using
|{1,2,3,...}| = sum_{i e N} 1 < sum_{i e N} i.

None of which is a mathematically valid proof.

I forgot even to mention the most basic proof:

From the three premises:
1) For all positive natural numbers n in N we have n > 0;
2) There is no negative natural number;
3) There is a natural number 1;
it follows that 1 + 2 + 3 + ..., if defined, cannot be less than 1.
Probably we can reduce the number of premises to two or even one. The
result stands irrevocable.

How can you overlook this simple conclusion?

Regards, WM

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