# How do we express the formal idea: The sum of even numbers?

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### wij

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Jan 19, 2022, 4:11:28 AMJan 19
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Rule3 of Infinite Series is added into NumberView-en.txt
https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download

Rule3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)
Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))

These rules are commonly used, but some issue may be controversial:

Ex: Σ 2*n =Σ (n+n) =Σ n + Σ n
If Σ 2*n is said the sum of all even numbers, Σ n the sum of all natural
numbers, the notion that the whole is greater than the part is conflicted
by this rule.
But, how do we express "the sum of even numbers"? Or Σ(n=0,∞/2) 2*n ?
An idea that using C-language's for loop expression might solve this
problem (or, at least, better than the traditional Σ notation):
for(n=0;;++n) n; or f(n=0;;n+=2) n;

Any idea?

### Malcolm McLean

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Jan 19, 2022, 5:24:42 AMJan 19
to
The sum of all the natural numbers is obvioulsy infinity.
So if we sum the even numbers, is that a lower infinity, because half
of the series has been removed? Or is it a higher infinity, because each
term is bigger?
It's quite hard to work out from first principles what the most logical
answers are.

### Ben Bacarisse

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Jan 19, 2022, 6:42:18 AMJan 19
to
wij <wyn...@gmail.com> writes:

> Rule3 of Infinite Series is added into NumberView-en.txt
> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
>
> Rule3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)
> Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))
>
> These rules are commonly used, but some issue may be controversial:
>
> Ex: Σ 2*n =Σ (n+n) =Σ n + Σ n

That's not an application of the rule you cite.

> If Σ 2*n is said the sum of all even numbers, Σ n the sum of all natural
> numbers, the notion that the whole is greater than the part is conflicted
> by this rule.

You will have a lot of trouble if you want mathematics to align with
your intuitions. You'd have to start by picking a definition of "greater
than" that matches your intuition and then you would have to follow
through with the consequences of that decision.

You could always just say there are no non-finite sets in your
mathematics. Do you need them? You could leave them just for people
who want them!

> But, how do we express "the sum of even numbers"?

Yes. Σ 2n is the sum of the even numbers.

--
Ben.

### Julio Di Egidio

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Jan 20, 2022, 4:02:33 AMJan 20
to
Hm, fantastic, another pile of Ben's steaming retarded shit.

*Spammer Alert*

Julio

### Richard Damon

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Jan 21, 2022, 10:43:54 AMJan 21
to
Fundamentally, the issue is that these rules are all under the condition
that the infinite sum converges to a finite value. When you try to think
of a unconverging sum as 'having a value', you can get all sorts of
issues. Even if the sum converges, there can be limits to what sort of
operations you can do on the elements if the sum has positive and
negative elements, and the sum of just the elements of one sign don't
also converge. Such sums can change their value if you reorder the terms.

Basically, 'infinity' doesn't act like a normal number, and you can't
expect it to, and infinite sums that don't converge fall into that category.

Ignoring that can get you into all sorts of problems, for example if we
let: (let sum(f(n)) be the sum of f(n) for all values of n from 0 to
infinity.)

X = sum(n)
Y = sum(1)

The sum of the odds can be expressed as either

sum(2n+1) = 2X+Y
or, if we skip the first element and add it in explicitly)
1 + sum(2n+3) = 1 + 2X + 3Y
thus it seems we could say that

2X+Y = 1 + 2X + 3Y

But this says that Y = -1/2 ( 0 = 1 + 2Y)

### wij

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Jan 22, 2022, 4:04:44 AMJan 22
to
On Friday, 21 January 2022 at 23:43:54 UTC+8, Richard Damon wrote:
> On 1/19/22 4:11 AM, wij wrote:
> > Rule3 of Infinite Series is added into NumberView-en.txt
> > https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
> >
> > Rule3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)
> > Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))
> >
> > These rules are commonly used, but some issue may be controversial:
> >
> > Ex: Σ 2*n =Σ (n+n) =Σ n + Σ n
> > If Σ 2*n is said the sum of all even numbers, Σ n the sum of all natural
> > numbers, the notion that the whole is greater than the part is conflicted
> > by this rule.
> > But, how do we express "the sum of even numbers"? Or Σ(n=0,∞/2) 2*n ?
> > An idea that using C-language's for loop expression might solve this
> > problem (or, at least, better than the traditional Σ notation):
> > for(n=0;;++n) n; or f(n=0;;n+=2) n;
> >
> > Any idea?
> Fundamentally, the issue is that these rules are all under the condition
> that the infinite sum converges to a finite value.

If the result is not known, how does one know the infinite series converges or not?
(converge OK, not converge not OK? what is the rule!)

> When you try to think
> of a unconverging sum as 'having a value', you can get all sorts of
> issues. Even if the sum converges, there can be limits to what sort of
> operations you can do on the elements if the sum has positive and
> negative elements, and the sum of just the elements of one sign don't
> also converge. Such sums can change their value if you reorder the terms.
>
> Basically, 'infinity' doesn't act like a normal number, and you can't
> expect it to, and infinite sums that don't converge fall into that category.
>
> Ignoring that can get you into all sorts of problems, for example if we
> let: (let sum(f(n)) be the sum of f(n) for all values of n from 0 to
> infinity.)
>
> X = sum(n)
> Y = sum(1)
>
> The sum of the odds can be expressed as either
>
> sum(2n+1) = 2X+Y
> or, if we skip the first element and add it in explicitly)
> 1 + sum(2n+3) = 1 + 2X + 3Y
> thus it seems we could say that
>
> 2X+Y = 1 + 2X + 3Y
>
> But this says that Y = -1/2 ( 0 = 1 + 2Y)

Lots of inconsistency bypassed for comment. I was asking for opinion, I won't say too much.

No valid rule for the derivation ∑ 2n*1= 1+ 2*(∑2*n+3) (trick of magic show)
A hidden false rule is assumed.

### Richard Damon

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Jan 22, 2022, 5:59:03 AMJan 22
to

On 1/22/22 4:04 AM, wij wrote:
> On Friday, 21 January 2022 at 23:43:54 UTC+8, Richard Damon wrote:
>> On 1/19/22 4:11 AM, wij wrote:
>>> Rule3 of Infinite Series is added into NumberView-en.txt
>>> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
>>>
>>> Rule3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)
>>> Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))
>>>
>>> These rules are commonly used, but some issue may be controversial:
>>>
>>> Ex: Σ 2*n =Σ (n+n) =Σ n + Σ n
>>> If Σ 2*n is said the sum of all even numbers, Σ n the sum of all natural
>>> numbers, the notion that the whole is greater than the part is conflicted
>>> by this rule.
>>> But, how do we express "the sum of even numbers"? Or Σ(n=0,∞/2) 2*n ?
>>> An idea that using C-language's for loop expression might solve this
>>> problem (or, at least, better than the traditional Σ notation):
>>> for(n=0;;++n) n; or f(n=0;;n+=2) n;
>>>
>>> Any idea?
>> Fundamentally, the issue is that these rules are all under the condition
>> that the infinite sum converges to a finite value.
>
> If the result is not known, how does one know the infinite series converges or not?
> (converge OK, not converge not OK? what is the rule!)

If the fact that the sum is finite is not known, you might not know if
the operation is valid, so you may need to carry the condition around in
the argument.

Just like if you have that X * Y = X * Z, you can't just divide by X and
say that Y = Z, you have to say something like if X != 0, then Y = Z.

>
>> When you try to think
>> of a unconverging sum as 'having a value', you can get all sorts of
>> issues. Even if the sum converges, there can be limits to what sort of
>> operations you can do on the elements if the sum has positive and
>> negative elements, and the sum of just the elements of one sign don't
>> also converge. Such sums can change their value if you reorder the terms.
>>
>> Basically, 'infinity' doesn't act like a normal number, and you can't
>> expect it to, and infinite sums that don't converge fall into that category.
>>
>> Ignoring that can get you into all sorts of problems, for example if we
>> let: (let sum(f(n)) be the sum of f(n) for all values of n from 0 to
>> infinity.)
>>
>> X = sum(n)
>> Y = sum(1)
>>
>> The sum of the odds can be expressed as either
>>
>> sum(2n+1) = 2X+Y
>> or, if we skip the first element and add it in explicitly)
>> 1 + sum(2n+3) = 1 + 2X + 3Y
>> thus it seems we could say that
>>
>> 2X+Y = 1 + 2X + 3Y
>>
>> But this says that Y = -1/2 ( 0 = 1 + 2Y)
>
> Lots of inconsistency bypassed for comment. I was asking for opinion, I won't say too much.
>
> No valid rule for the derivation ∑ 2n*1= 1+ 2*(∑2*n+3) (trick of magic show)
> A hidden false rule is assumed.

The only hidden false rule that is assumed is that you can manipulate
infinite sums like normal number, even if they don't converge.
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