Prove me wrong

That's not proof. Amd the point about infinity is that you don't stop.

1-1+1-1+1-1+1-... = 1 - (1-1) - (1-1) - (1-1) - ...

= 1 - 0 - 0 - 0 - ... = 1

Infinite repetition

>Posting a precalculus homework question
Still stuck on sums of infinite series huh?

so, like a shitpost?

The series is not a Cauchy series therefore it does not converge anywhere in the real numbers.

Q.E.D.

... = -1 / 12

Adding or subtracting two or more integers will always result in an integer. 1/2 is not an integer. Git gud.

Let S = 1 - 1 + 1 - 1 + ...
1 - S = 1 - (1 - 1 + 1 - 1 + ...) = 1 - 1 + 1 - 1 + ... = S
1 - S = S
1 = 2*S
S = 1/2
1 - 1 + 1 - 1 + ... = 1/2

LOL get rekt noob

No, it will either be 0 or 1

I have no doubt people believe this prooves it.

[math]1\not\to 0[/math]

wow ebin math :DDD

S = 1 + 1 + 1 + 1 + ...
1 + S = 1 + (1 + 1 + 1 + ...) = S
S = S + 1
0 = 1

broof that 0 = 1 :-----D

So I guess we can conclude that 0 = 1/2 = 1

[math] |z|

Direct comparison test, harmonic numbers, q.e.d.

Damn I really want to try that. Also checked

it might be

It depends when you cut off.

Lol math does crazy stupid hocus pocus 24/7, and the second people can't wrap their mind around it they suddenly become experts, calling bullshit on basic operations. It's hilarious. I had one kid like this in my algebra 2 in freshman year he was such a brainlet that he was trying to argue against i and e, what a cuck

I don't like the whole "it's where you place the brackets" thing like when the fuck did people decide you could do shit arbitrarily and not left to right and in the order of operations. It should be 0.

winner

sage

where is the error in this

he concluded that 1 + (1 + 1 + 1 + ...) = S, which lacks proof

But if you put the brackets like this
1-1+1-1+1-1+...
= (1-1)+(1-1)+(1-1)+...
= 0+0+0+... = 0

Is this impossible to prove?

Because adding one to an infinite of series of ones equals one yeah?

If you were to take the limit of (x+1)/x as x approaches infinity you get one
Unless you're a 1.0000.....1 =/= 1 faggot

if [math](1-1)+(1-1)+\ldots=0[/math] and [math]1+(-1+1)+(-1+1)+\ldots=1[/math] then [math]1-1+(1-1)+(1-1)+\ldots=2[/math] and so on right?

equals an infinite series of ones*
lol

Hm, how're the Numberphiles videos? Care to actually think what any of that means?

When you change the sequence of an infinite series, you change its sum.

That series doesn't converge uniformly dumbass

You can come up with all kinds of nonsense when working with a divergent series. It's why we leave and undefined.

This is correct. There is not proving you wrong.
It's 1/2 because, essentially, as you go to infinity, you're making an infinite average of zeros and ones, and the average of those two happens to be, surprise surprise, 1/2.
That's a simple explanation without the rigour of limits and series.

...

So it could be like an infinite sum of :
1+(i).(i)

wtf, there was nothing wrong with my geometric series

[math] \lim_{\epsilon \to 0} \sum_{n=0}^\infty (-1)^n \cdot (1 - \epsilon)^n = \dfrac{1}{2} [/math]

Can be proven in analysis.

However
[math] \lim_{\epsilon \to 0} \sum_{n=0}^\infty (-1)^n \cdot (1 - \epsilon)^n \ne \sum_{n=0}^\infty (-1)^n \cdot \lim_{\epsilon \to 0} (1 - \epsilon)^n = \sum_{n=0}^\infty (-1)^n [/math]

The two possible answers are 0 and 1 depending on when you stop so the average of 0 and 1 is 1/2. It's not that hard, moron.

1-1+1-1+1-1...... = [0 ; 1]

Functions with more than two possible outputs, mutually exclusive to each other, aren't uncommon

Whoever can't understand that should get shot

>1 - 1 = 0
>Replace every instance of 1 - 1 with 0
>0 + 0 = 0
>0 + 0 + 0 + 0... = 0

you're never cutting off, thats the quesiton

It's literally not a function in that case.

The dots mean +0.5

>Function.
>Having 2 ranges to a single domain.

whoever says it equals to any number is a brainlet

it is an alternating series.
take the subsequence a to be 1, 1, 1, 1,...
take the subsequence b to be -1, -1, -1, -1,...
take the limit to infinity of n for a and b. they are not equal therefore they diverge.

a sequence converges if those two limits are equal.

now fuck off with this thread

The thread should have ended after this

If infinity is even, then 1-1+1-1+1-1...=1
If infinity is odd, then 1-1+1-1+1-1...=0
If infinity is neither, then 1-1+1-1+1-1...=undefined
If infinity is both, then 1-1+1-1+1-1...=undefined
I'm pretty sure infinity is neither/both, so the sum doesn't exist.

An infinite sum converges if the sequence of it's partial sums tends to a limit. So in that case we get {0,1,0,1,...}. This sequences doesn't converge and so doesn't the series.