1-1+1-1+1-1+1...? #shorts

This series goes on forever. Sometimes it's 1. Sometimes it's 0. Can it ever be 1/2? Can we even know? WHAT IS THE ANSWER? #shorts

[aaronmihaljevich]

I don't come here for questions, I come here for answers man!

[spodefollower]

It’s an infinite series that doesn’t converge. So there’s no answer

Edit: To everyone saying it’s “0 or 1”. Yes it oscillates between those values, but it is an *infinite* series. The nature of infinity is such that it cannot be “0 or 1”, as you are imagining the series stopping or terminating at some point, (or looking at some individual point within the series) at which the final value will be 0 or 1. But it does not stop, ever, so it will never be either nor both of those values. It is literally impossible to define this series by any value (ramanujan summation excluded, as that isn’t really a true “sum” of the series, as the average person would understand a sum to be). I know it isn’t satisfying and goes against your intuition but there truly isn’t an answer in the classical sense.

[UnboxJunkie]

If we're considering the moments in time, it's either 0 or 1 depending on which step you're on at that moment. If time was finite and we're looking at its entire existence, its both 0 and 1.

[vVearon]

i spit my water out when he said *”2000”*

[paIapin]

I was so scared when he said "2000"


but then he said "students"

[pheto]

"What is this'"
*Doesn't tell us what it is*
Me: He's legit asking us because he don't know....

[wyboo]

for everyone that wants a serious mathematical answer:

it depends on your definition of infinite summation. the regular definition of an infinite sum is the limit of the partial sums. for example, consider the infinite sum:

1 + 1/2 + 1/4 + 1/8 + 1/16 + ...

where each term is half of the previous one. the usual way we do infinite sums is by taking partial sums and seeing what they approach:

1
1 + 1/2 = 3/2
1 + 1/2 + 1/4 = 7/4
1 + 1/2 + 1/4 + 1/8 = 15/8

We can see that the partial sum with the first n terms is:

(2^n-1)/(2^(n-1))

as n approaches infinity, the -1 becomes negligent compared to the exponential (you can do this with limits), so this infinite sum equals (2^n)/(2^(n-1))=2^(n-n+1)=2

if we use this definition of infinite summation, the series in this short doesn't converge because the partial sums just bounce back and forth between 0 and 1:

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

BUT, there's no reason to say that this definition of infinite summation is the only correct one. in particular, there's an equally valid definition of infinite summation where the value of an infinite sum is the limit of the AVERAGE of its partial sums rather than the limit of its partial sums directly. this is a real definition mathematicians use that has real actually applications; it's called Cesaro summation. applied to this series, which is known to mathematicians as Grandi's Series, we can assign a perfectly sensible value.

The nth partial sum of Grandi's Series is 1 when n is odd and 0 when n is odd. so the Cesaro summation of Grandi's series will be the average of the first n partial sums as n goes to infinity:

partial sums: 1, 0, 1, 0, ...
average of first n partial sums:

(1+0+1+0+...)/n

in the numerator, we have basically half ones and half 0s in the limit, so the numerator tends to n/2, so:

(n/2)/n=n/(2n)=1/2

so the Cesaro summation of Grandi's Series is 1/2

[visibleshow]

-If it begins with a minus and ends with a plus or the opposite it is: 1
-if it begins with a minus and ends with a minus and ends with minus it is: 0
- if it begins with a plus and ends with a plus it is: 2.

[eximius]

Me: I'm just waiting for an equal sign, then I'll answer it.

[-.._.-._.-_--._---_-.]

1+1–1-1–1+1- is sometimes called Grandi's series.
One obvious method to attack the series
1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + ...

is to treat it like a telescoping series and perform the subtractions in place:

(1 − 1) + (1 − 1) + (1 − 1) + ... = 0 + 0 + 0 + ... = 0.

On the other hand, a similar bracketing procedure leads to the apparently contradictory result

1 + (−1 + 1) + (−1 + 1) + (−1 + 1) + ... = 1 + 0 + 0 + 0 + ... = 1.

Thus, by applying parentheses to Grandi's series in different ways, one can obtain either 0 or 1 as a "value".

Treating Grandi's series as a divergent geometric series, we may use the same algebraic methods that evaluate convergent geometric series to obtain a third value:

S = 1 − 1 + 1 − 1 + ..., so

1 − S = 1 − (1 − 1 + 1 − 1 + ...) = 1 − 1 + 1 − 1 + ... = S

1 - S = S

1 = 2S,

resulting in S = 1/2. The same conclusion results from calculating −S, subtracting the result from S, and solving 2S = 1.

Therefore, one can arrive at two conclusions:

The series 1 − 1 + 1 − 1 + ... has no sum.
...but its sum should be 1/2.
Hope This Helps!!!

[Pixelcraftian]

I didn’t understand at first so I thought the pluses and minuses were jailed for their crimes

[chimiseanga]

There are essentially two ways to attack this problem:
- Classically, this is an infinite series which diverges since the sequence of partial sums oscillates between 0 and 1 forever, so the sum has no answer
- Using Cesaro sums, whereby you take the limit of the arithmetic means of the first n partial sums as n goes to infinity, you get 1/2. This is a different interpretation of infinite sums, but it has some credence since the Cesaro sum of a converging series (in the classical sense) is equal to its usual classical sum.

[Kahandran]

Kevin: *snorts a line of coke*
"YEAH LET'S MAKE A SHORT"

[danielle-]

As long as he did not say "equal" then theres no answer

[jonatanskulpe]

Bro, the 1-1+1-1+1-1+1 turned into a sick beat

[Anime-yzfk]

The sum of an infinite series is defined to be the limit of the sequence of its partial sums, if it exists. The sequence of partial sums of above series i.e Grandi's series is 1, 0, 1, 0, ..., which clearly does not approach any number (although it does have two accumulation points at 0 and 1). Therefore, Grandi's series is divergent and its sum cannot have a definite answer

That's the explanation which I copied from the net.

[Swayto-]

No answer because this is technically an expression, not an equation

[CenterBloc]

Depends if it ends with a + then its 1 if it ends with a - then its 0
Explanation 1-1=0+1=1 so and so forth until it reaches + or - if its well - thene 0 but plus is 1

[alapansen]

I seriously didn't even noticed those 1s as they were looking like modulus signs 😂😂😂😂

[jimmyraymond]

its depends on how fast you can count in 10 mins.