So long stackers 😈
Identify the catch and get rewarded!
We know
and
(subtracting the number from itself)
Thus
[Grandi's Series]
Hence
Thus
Or
[Leave the first term and take the "-" common]
Thus
SO
ha 😈
1,000 sats paid
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noknees's bounties
The catch is in the 6th step, where the extraction of the (-) is made so that it stealthy breaks the assumed parity of the series:
in 1 - (1 - 1 + 1 - 1 + ... ) there's a hidden (+ 1) at the end, so the series should actually look like:
1 - (1 - 1 + 1 - 1 + ... + 1 - 1 + 1)
so close!
but the explanation is not quite right
I could express it like so:
The proposition starts with ( 1 - 1 ) + ( 1 - 1 ) + ( 1 - 1 ) = 0
Then expands it, but it does so ignoring the last part. We can make the same infinite expansion keeping the representation of the last part, like so:
( 1 - 1 ) + ( 1 - 1 ) + ... + ( 1 - 1 ) = 0 , instead of ( 1 - 1 ) + ( 1 - 1 ) + ... = 0
If we take the (-1) from the new representation, we can see that in reality what we get is:
1 - (1 - 1 + 1 - 1 + ... + 1 - 1 + 1) = 0
while the 1 - (1 - 1 + 1 - 1 + ... ) = 0 notation makes it look like if the series ends, in the infinite, like
1 - (1 - 1 + 1 - 1 + ... + 1 - 1 ) = 0
Which is not the true form of the first proposed series.
nailed it!
it's in the number of terms!
Awesome! Thank you for making this fun and for the bounty, much appreciated :)
You played well.
Finally :')
(1-1)+(1-1)+(1-1)+... = 0
1-((1-1)+(1-1)+(1-1)+...+1) = 0
0 = 0
nice try, fed
you're late but here's a consolation
1−(1−1+1−1+…)=1 [Leave the first term and take the "-" common]
Thus
1=1+(1−1+1−1+1−1....)
nope, no new term was introduced, the negative sign is just taken as common leaving the first term as is
1−(1−1+1−1+…)=0
Isn't this equation wrong?
How?
Just appearing on the equation at first glance will make that seem wrong if you don't follow the previous steps :)
That's not the breaking point
1−(0)=0
let me explain,
it was
1 - 1 + 1 - 1 + 1 - 1.....
So
I kept the first 1, then took the negative sign common for the rest
eg
8 - 3 - 5 = 0
or 8 - (3 + 5) = 0
It might look illusionary lol
My head's spinning! Ahahah
😂
I looked through the comments to see if someone mentioned the order of operation. You could say it is this at a first glance, but... nope.
False proofs always appear because they are generally presented with what seems to be valid math as in the example above.
The mistake in this case lies in treating the infinite series 1-1+1-1+1-1+....= 0.
This is the catch.
Grandi’s series does not converge in the traditional sense - its partial sums just alternate between 1 and 0. You are basically manipulating the result of this and force it to 0 artificially.
This is my opinion.
that's the point of this trick!
1−(1−1+1−1+…)=0 -> this is the moment of failure.
You are implying Grandi's series is =1 when you said before it =0.
well, you are saying this #1008494
too close! too close!
just one subtraction close!
Interesting. I've usually seen Grandi's series as 1-1+1-1+1-1+1-1...
That series doesn't converge to 0 by any accepted method I know. Infinite series are weird like that!
it is a divergent series and is still divergent!
you can get 2 answers out of this :)
hmmm...when you move first 1 left of equal sign you have one term left and so 5 terms right ( considering the example of 3 couples so 6 ones ). So in the end:
1 = 1-1+1-1+1
so 1 = 1
no! I'm not moving the ones anywhere :)
I just took the minus common and the rest of the series to the other side
but you're close, try again
@south_korea_ln is not allowed to answer this :)
Cesàro thinks your logic also implies that 1/2 = 0...
😂 that's the OG series ofc why not