Math: Addition "factorial"???

[SpaceBadger]

I'm sure there must be a term for this, but if I ever learned it I have forgotten.

I remember factorials for multiplication. Four factorial (4!) = 4x3x2x1 = 24.

What is the equivalent term for addition? 4+3+2+1 = 10. What is the term for that? Is there a standard notation, as in using ! for factorials?

This is in connection with a homebrew, but didn't seem appropriate to post in that subforum, so I hope this is the right place to ask.

[Heliomance]

No, pretty sure there isn't. The only option is Σⁿ_i=1i

[Jimorian]

Just "Sum". Or if you need to specify the range or other limitations "Sum from 4 to 8", "Sum of even numbers from 42 to 116".

[Tirian]

As far as a term for them goes, they are called triangular numbers. You'll see people saying things like "The fifth triangular number is 15", so that has some intuitive and semi-common meaning.

It isn't really useful enough to deserve a concise notation like factorial has. Also, it's really just n(n+1)/2, so I figure most people just write it like that.

But as long as you are upfront and consistent with your notation, you've have flexibility to create notation in your own context. For instance if you want to say 7△=28, knock yourself out.

[Anxe]

They're called triangle numbers because of this:

1
22
333
4444
55555

Add up one number from each tier of the right angled triangle there and you get the 5th triangle number, 15. And as Tirian said, the algebraic formula is used most often for it, but feel free to invent your own. Perhaps !5 could be 15, so that we preserve a little bit of that factorial love.

[SpaceBadger]

Thanks, y'all!

[Riverdance]

No, pretty sure there isn't. The only option is Σⁿ_i=1i

That Greek E symbol is called Sigma. It's like the addition version of factorials.

[The Extinguisher]

No, pretty sure there isn't. The only option is Σⁿ_i=1i

This is the best way to do it, because it doesn't involve any new notation and everyone will understand what you're talking about.

Plus, it allows you to take the sum of just even numbers, or numbers 7 through 89, or whatever else you want.

[Riverdance]

There are two categories of series. Convergent and divergent. Divergent series are simple. They diverge to infinity, and example being 1+2+3+4+5+....

Convergent series are not so simple. They converge to a particular number, that is, instead of approaching infinity, they come infinitely close to a finite value. An example is 1/3+1/9+1/27+1/81+.... This can be written in sigma notation as Σ 1/(3^k) with k starting at 1. This series is part of a group called "geometric series." This particular one converges to 1/2. It's difficult to explain this with the limited graphics of a forum posting. I recommend looking it up or finding someone who knows basic calculus.

Edit: Heliomance, how did you get the sigma symbol into your post? I had to copy and paste it from yours. Did you just copy and paste it from somewhere else?

[Mewtarthio]

That Greek E symbol is called Sigma. It's like the addition version of factorials.

Strictly speaking, it's the addition version of Capital Pi. Now, if you swap out the Sigma in Heliomance's post with a Capital Pi, you get the factorial, so it's sort of close.

[Heliomance]

Edit: Heliomance, how did you get the sigma symbol into your post? I had to copy and paste it from yours. Did you just copy and paste it from somewhere else?

Character Map. Apparently Alt+228 is supposed to do it as well, but I get õ instead.

[Kalirren]

Yeah. There isn't really special notation for it because it's really easy to evaluate that sum:

n+(n-1)+...+2+1 = n(n+1)/2

[danzibr]

There are two categories of series. Convergent and divergent. Divergent series are simple. They diverge to infinity, and example being 1+2+3+4+5+....

Not true. You can have divergent series which don't diverge to plus or minus infinity. An example being 1-1+1-1+1-1+1-1+...

Anyways, even the factorial symbol is just a specific case of using the capital pi notation.

[PirateMonk]

Convergent series are not so simple. They converge to a particular number, that is, instead of approaching infinity, they come infinitely close to a finite value. An example is 1/3+1/9+1/27+1/81+.... This can be written in sigma notation as Σ 1/(3^k) with k starting at 1. This series is part of a group called "geometric series." This particular one converges to 1/2. It's difficult to explain this with the limited graphics of a forum posting. I recommend looking it up or finding someone who knows basic calculus.

Not really, at least for geometric series. Paraphrased from Wikipedia:

Take a geometric series with first term a and common ratio r
Let s=a+ar+ar²+...+ar^n-1
Therefore, s-rs=a-arⁿ
Factor: s(1-r)=a(1-rⁿ)
Divide by (1-r): s=a(1-rⁿ)/(1-r)
For |r|<1, rⁿ gets increasingly close to zero as n grows (i.e. as you add increasing many terms). This step is technically basic calculus, but should be understandable to anyone with a reasonable grasp of algebra.
Therefore, if each term of a geometric series is closer to zero than the previous one, the infinite sum is a/(1-r)
In the example above, a=1/3=r. 1-1/3=2/3. (1/3)/(2/3)=1/2.

[Dogmantra]

Character Map. Apparently Alt+228 is supposed to do it as well, but I get õ instead.

I would also recommend copypastecharacter.com but only because I am in love with that website.

[Tirian]

I just Google "sigma" or any mathematical symbol or foreign letter. The top hit will be for the wikipedia page and the Unicode expression will be in the description. Copy it into your clipboard and off you go.

[Knaight]

Not really, at least for geometric series. Paraphrased from Wikipedia:

Geometric series represent only a tiny fraction of convergent series. For instance Σ1/ln(k) starting at 1 is convergent (though I didn't double check this), but the infinite geometric series formula is completely worthless, and calculus is actually necessary. In any case, regarding the original question the term is "arithmetic series", though 1+2+3+4....+(n-1)+(n) is only one of them.

[Razanir]

Not true. You can have divergent series which don't diverge to plus or minus infinity. An example being 1-1+1-1+1-1+1-1+...

That one doesn't actually diverge. The infinite sum can be expressed as the partial sum from 1 to n as n approaches infinity. In your example, the limit simply does not exist. That is different from diverging.

As far as a term for them goes, they are called triangular numbers. You'll see people saying things like "The fifth triangular number is 15", so that has some intuitive and semi-common meaning.

It isn't really useful enough to deserve a concise notation like factorial has. Also, it's really just n(n+1)/2, so I figure most people just write it like that.

But as long as you are upfront and consistent with your notation, you've have flexibility to create notation in your own context. For instance if you want to say 7△=28, knock yourself out.

I've always called them Gaussian Sums, myself

[John Cribati]

It's officially called "Summation from x to y" by the way.

[PirateMonk]

Geometric series represent only a tiny fraction of convergent series. For instance Σ1/ln(k) starting at 1 is convergent (though I didn't double check this), but the infinite geometric series formula is completely worthless, and calculus is actually necessary. In any case, regarding the original question the term is "arithmetic series", though 1+2+3+4....+(n-1)+(n) is only one of them.

I assumed that Riverdance used "this" refer to the convergence of a particular geometric series. If it was intended to mean convergence in general, then you would need calculus.

Also, Σ1/k diverges, and x>lnx, so I would be very surprised if Σ1/lnk converged.

[Razanir]

Also, Σ1/k diverges, and x>lnx, so I would be very surprised if Σ1/lnk converged.

Your answer is right, but the logic is wrong.

Direct Comparison Test (what you tried):
If 0<a<b for all n,
If b converges, so does a. And if a diverges, so does b

Nth Term:
lim(n -> infinity) [1/(ln n)]
1/(ln infinity) = 1/infinity = 0
Because the limit equals 0, the sum converges

[The Extinguisher]

Your answer is right, but the logic is wrong.

Direct Comparison Test (what you tried):
If 0<a<b for all n,
If b converges, so does a. And if a diverges, so does b

Nth Term:
lim(n -> infinity) [1/(ln n)]
1/(ln infinity) = 1/infinity = 0
Because the limit equals 0, the sum converges

That's what he did.

x > lnx is the same as 1/x < 1/lnx
Both are greater than zero, so the direct comparison works.

And your second method doesn't confirm that it converges. All it does is say that it might converge. (or rather, that the series itself converges) You need another method to see if the partial sums converge.

[Razanir]

x > lnx is the same as 1/x < 1/lnx
Both are greater than zero, so the direct comparison works.

And your second method doesn't confirm that it converges. All it does is say that it might converge. (or rather, that the series itself converges) You need another method to see if the partial sums converge.

I see what I did wrong. I confused all my signs then tried to argue the wrong point. I confirmed it with the Integral Test (easier for me because it's more math and less conceptual):

integral(2 to infinity) [1/(ln n) dn]
integration by parts-
u = 1/(ln n)
du = 1/(n * ln n * ln n) dn
dv = dn
v = n

The first term is n/(ln n)

Plug in infinity, it's infinity/infinity. L'Hôpital gives 1/(1/n) = n = infinity

Yep, it diverges! My bad

[danzibr]

That one doesn't actually diverge. The infinite sum can be expressed as the partial sum from 1 to n as n approaches infinity. In your example, the limit simply does not exist. That is different from diverging.

Then we are using different definition. For me, a series converges to a real number, say L, if for every positive epsilon there exists a natural number N blah blah blah.

It diverges, by definition, if it does not converge. So yes, the series I was describing diverges. It does *not* diverge to plus or minus infinity, but it does diverge.

[danzibr]

That's what he did.

x > lnx is the same as 1/x < 1/lnx
Both are greater than zero, so the direct comparison works.

And your second method doesn't confirm that it converges. All it does is say that it might converge. (or rather, that the series itself converges) You need another method to see if the partial sums converge.

This is the correct math btw. No need for integral test.

[Razanir]

Yeah, but if the integral's not too hard, the integral test is fun! I like getting to plug infinity into functions

[Heliomance]

Then we are using different definition. For me, a series converges to a real number, say L, if for every positive epsilon there exists a natural number N blah blah blah.

It diverges, by definition, if it does not converge. So yes, the series I was describing diverges. It does *not* diverge to plus or minus infinity, but it does diverge.

This is correct. Divergent series need not diverge to plus/minus infinity, it is sufficient that they do not converge to a finite limit.

[danzibr]

This is correct. Divergent series need not diverge to plus/minus infinity, it is sufficient that they do not converge to a finite limit.

Thanks. I was wondering if such a basic definition would actually differ from text to text.

[ZyChrome-Inc]

I'm sure there must be a term for this, but if I ever learned it I have forgotten.

I remember factorials for multiplication. Four factorial (4!) = 4x3x2x1 = 24.

What is the equivalent term for addition? 4+3+2+1 = 10. What is the term for that? Is there a standard notation, as in using ! for factorials?

This is in connection with a homebrew, but didn't seem appropriate to post in that subforum, so I hope this is the right place to ask.

Hello! (i joined solely for this) i have the same issue, and have been recently using this notation, (with explanation)

n(+!)=n+(n-1)+...+2+1
so 4(+!)=10; 3(+!)=6

note on paper: 'additive factorial, add all smaller whole positive numbers.'

the full definition: +! additive-factorial, factorial style, with + instead of *

[Jay R]

Yes, if a series does not converge to a single number, it diverges. I have not seen a math text that disagrees.

For instance Σ1/ln(k) starting at 1 ...

"starting at 1"?

Am I the only one who noticed that the first term is undefined?

Hello! (i joined solely for this) i have the same issue, and have been recently using this notation, (with explanation)

n(+!)=n+(n-1)+...+2+1
so 4(+!)=10; 3(+!)=6

note on paper: 'additive factorial, add all smaller whole positive numbers.'

the full definition: +! additive-factorial, factorial style, with + instead of *

Since they are triangular numbers, on the rare occasions I need them, I refer to them as △(n) or △_n, or just use the formula n(n+1)/2.