How the fuck is the series 1-1+1-1+1-1.... divergent?

How the fuck is the series 1-1+1-1+1-1.... divergent?

It literally sums to zero, even a dumbass could tell from one glance that the series is convergent.


infinite positive one + infinite negative one = zero

Other urls found in this thread:

youtube.com/watch?v=-EtHF5ND3_s
symbolab.com/solver/step-by-step/\sum_{n=0}^{\infty} n/2^{n}
en.wikipedia.org/wiki/Riemann_series_theorem
twitter.com/SFWRedditVideos

>even a dumbass could tell from one glance that the series is convergent.
Then give a proof.

The sequence of partial sums does not converge, therefore it is not convergent.

easy

Let A = B

therefore A - B = 0

thus A - B + A - B.....n = n(A) - n(B) = n(A-B) = 0


literally middleschool tier.

To converge to 0 means that there exists a number N such that after N terms in the series, the difference between 0 and the series is less than any number you want to choose.

We say it is divergent because we have not found a number N such that the difference between 0 and the series is less than any number you choose.

>assuming what you're trying to prove in the third step
literally middleschool mistake

For the same reason you can't assign a value to
[math]\sin(\infty)[/math]

not op, but what happens when you group them in twos and reform the set, reducing the number length to n/2?

You can group the "adjacent" elements of the sum in multiple ways and get different answers. Consider the following two ways:

i) S = (1-1)+(1-1)+...
ii) S = 1 - (1-1) -(1-1) - ...

If you believe this procedure is valid, then in case i) we arrive at S=0, where as in case ii) we arrive at the conclusion S=1.

The actual conclusion we have to make is that a formally divergent sum cannot be consistently partitioned into separate sums, i.e. the procedure is invalid.

Ah I see. Thanks.

Divergent means that the value of a series cannot be "estimated".

No problem. I recommend checking out the youtube channel Mathologer. He has formal training in mathematics and likes to talk about these things and explain them in a really precise manner.

youtube.com/watch?v=-EtHF5ND3_s

This is why maths is brain masterbation.
Engineering and physics are better.

You get different values for its convergence dependinv on how you arrange the terms. Do you not understand infinity???

Can't remember the details of it but I watched a numberpedo video about the -1/12 meme and part of the proof for that rode on assuming the value of this series is 1/2, so don't do that.

EVERYONE IS WRONG IT CONVERGES TO HALF

PROOF:

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

think about it this way: if you have a switch turned off at t=0 and you turn it on at t=1/2, off at t=3/4, on at t=7/8, off at t=15/16, etc. what will the state of the switch be at t=1? it will be oscillating between on and off so fast that it can only be considered both on and off. if on is 1 and off is 0, the switch is at 1/2.

This proof is bad and you should feel bad.

1-1+1-1+1...
=1+(-1+1)+(-1+1)+(-1+1)...
=1+0+0+0+0...=1

which is it pal?

It is neither, for this sequence diverges.

oh shit I took this guy's undergrad math class a couple of years ago

You all are (wrongly) assuming that the sum has an even number of terms.

Since the amount of terms is infinite, we can't know whether it has an even or odd amount of terms.

Diverging to infinity isn't the only way to diverge.

The partial sums alternate between 1 and 0.

divergent doesn't equal "not convergent"
sin is a periodic function, so it has no limit but that doesn-t mean it is divergent as its value is bounded (by 1 and -1)

infiniti is not a number, is not like 88 or 4, 8, 69 etc.

(3^n/2^n) when n go to in infiniti. S we have infiniti/infiniti. and it is 1 for You? Take a random smal n like 55.

For n = 55 we have 3^55/2^55 and for calculator this is 4841938267.25 not 1, not infiniti/infiniti not 2/2, not 55/55. so infiniti =/= infiniti

So infiniti + infiniti =/= 2 infiniti
infiniti + infiniti = infiniti
So infiniti - infiniti = ? We cannot do this.

Other example(n go to infiniti):

5^n-3^n = ?
>infinite positive one + infinite negative one = zero
say 0.


5^n-3^n = ?
I say this is infiniti, because 5 > 3.
5-3=2
25-9=16
125-27=98
...

>How the fuck is the series 1-1+1-1+1-1.... divergent?

Simple that series not going to 1 or -1, or 0. Sometimes is 1 sometimes is 0. We can't say (-1)^n when n go to infiniti and n start from 2, he go to a one number.

Example: n go to infiniti, n >0

E(1/n^2)

1/1+1/4+1/9+1/16+1/25+1/36+1/49+1/64+1/81+1/100

(carefull, 1/n is divergent)

Other example: symbolab.com/solver/step-by-step/\sum_{n=0}^{\infty} n/2^{n}

You canno rearrarange the summands if the series doesn't converge absolutely.

en.wikipedia.org/wiki/Riemann_series_theorem

How the fuck is the series 1-1+1-1+1-1.... divergent?

It literally sums to one, even a dumbass could tell from one glance that the series is convergent.

1-1+1-1+1-1.... = 1-(1-1+1-1+1-1...) = 1 - 0 = 1

one + infinite positive one + infinite negative one = one

Take the terms in the positions 1,3,5,... That summatory gives infinite positive.
Take the terms in the positions 2,4,6,... and gives infinite negative.
Then, it is imposible that the whole sum results on a single number. QED.

Is infinity odd or even? As you can see, the partial sums are 1 for odd and 0 for even numbers. What you are claiming is that infinity is even, but you can write [math]\infty = 2\infty +1[/math], so it is also odd. The converse holds too: if infinity is odd, then it is even. Therefore you also get 1 as the sum. Can you see my point, OP?

>masterbation
Physics is a good choice for you, barbarian.