If you keep halving something, in theory you never reach zero. Between any two numbers on the real number line there are an infinite number of other numbers....it isn't amazing, it's banal.
Gibzie, The nth term is 2. It is a finite number.
Oh, and by the way, 1.999... = 2, so your answer is also correct. You may want to check out the 1.999... = 2 topic on wikipedia or other math-related places.
i dis agree with you m8, i figer: the n'th term no matter how infinate will never reach 2, because if it could reach 2 then it woud be a finite nomber, as you have a goal... the point is to be gramaticly corect you would have to put that 1.99 reocuring would be your answar.... thanx
If you mean that if you could walk along the number line and actually step on each number 9 of 1.999999. . .(as 9 repeats infinitely) you would never step onto or "reach" 2.0. Notwithstanding, 1.999... EQUALS 2.0.
Dear Sir, What is the sum of the infinite series: 1-1/2+1/4-1/8+1/16 .. etc. I would like to be able to cut out the haggling process when purchasing items in a market... ;o)
Sum to infinity = a/(1-r), and in your case r = -(1/2), and a = 1, so you get 1/(1-(-1/2)) = 1/(3/2) = 2/3. So as long as you can explain this to the shopkeeper, you'll save a lot of time.
I don't get it so as you continue to ad how could this ever be 2, if it gives 2 then you could continue and it would give more than 2 right? so I don't see how this could get up to 2 it will always continue to be closer to 2 as you get further in the series
Sum{n=1; inf; 1/(2*n)} is said to converge to 2. This idea is the foundation of Calculus. If you can prove it to be false, then I you'd be in line for a Nobel Prize.
@mathproblems no its not... its @jlsdev is right... its infinitely bit less than 2.... plus, u need to pronounce INFINITE correctly.. not (INFIIIINIIIIIIITE...!)
This has been flagged as spam show
At the very end of your formula.... {2^(n) -1 / 2^(n-1)}
How did the numerator 2^(n) -1 become 2 whole minus 1 over the denominator??
robobrain10000 9 months ago
Comment removed
robobrain10000 9 months ago
Thanks i needed this, my teacher in my school doesent give us lessons she makes us learn them by reading
Juancamy 2 years ago
Comment removed
Srdjanr271 2 years ago
u could use the formulas a1/1-r and lim as n approaches infinite of 1/2 to the N+1 power over 1/2-1 = 0-1/1/2-1= -1/1/2-1=1/1-1/2= 2
turbold1 3 years ago
If you keep halving something, in theory you never reach zero. Between any two numbers on the real number line there are an infinite number of other numbers....it isn't amazing, it's banal.
Quietlydoesit 4 years ago
thanks for posting =D
devarkk 5 years ago
Gibzie, The nth term is 2. It is a finite number.
Oh, and by the way, 1.999... = 2, so your answer is also correct. You may want to check out the 1.999... = 2 topic on wikipedia or other math-related places.
EternalHarmonics 5 years ago
i dis agree with you m8, i figer: the n'th term no matter how infinate will never reach 2, because if it could reach 2 then it woud be a finite nomber, as you have a goal... the point is to be gramaticly corect you would have to put that 1.99 reocuring would be your answar.... thanx
gibzie 5 years ago
If you mean that if you could walk along the number line and actually step on each number 9 of 1.999999. . .(as 9 repeats infinitely) you would never step onto or "reach" 2.0. Notwithstanding, 1.999... EQUALS 2.0.
InfinitelyManic 3 years ago
Dear Sir, What is the sum of the infinite series: 1-1/2+1/4-1/8+1/16 .. etc. I would like to be able to cut out the haggling process when purchasing items in a market... ;o)
saiello2061 5 years ago
Sum to infinity = a/(1-r), and in your case r = -(1/2), and a = 1, so you get 1/(1-(-1/2)) = 1/(3/2) = 2/3. So as long as you can explain this to the shopkeeper, you'll save a lot of time.
mrmoo2001 5 years ago
this is beyond my intelligence... props to you.
sarielaxx 5 years ago
An infinitely tiny bit less than 2
jlsdev 5 years ago 2
it is 2
mathproblems 5 years ago
I don't get it so as you continue to ad how could this ever be 2, if it gives 2 then you could continue and it would give more than 2 right? so I don't see how this could get up to 2 it will always continue to be closer to 2 as you get further in the series
TellusJD 5 years ago
Thats 1+1/2+1/4+... approaches 2. But I wanted to know 1-(minus)1/2+1/4-(minus)1/8... which should be somwhere around 2/3, no?
Thanks.
saiello2061 5 years ago
That's where the term "convergence" comes in.
Sum{n=1; inf; 1/(2*n)} is said to converge to 2. This idea is the foundation of Calculus. If you can prove it to be false, then I you'd be in line for a Nobel Prize.
jmckaskle 4 years ago
Correction: Sum{n=0; inf; (1/2)^n}
Sorry :)
jmckaskle 4 years ago
That will still converge to 2. I want to know: 1 -(minus) 1/2 +(plus) 1/4 -(minus) 1/8 +(plus) 1/16 ... this should converge on the value of 2/3.
saiello2061 4 years ago
@jmckaskle
{n=1; inf; 1/(2*n)} converges to 1. I think you meant n=0
Paucharka 1 year ago
@jmckaskle
you already corrected it, sorry )
Paucharka 1 year ago
@mathproblems no its not... its @jlsdev is right... its infinitely bit less than 2.... plus, u need to pronounce INFINITE correctly.. not (INFIIIINIIIIIIITE...!)
hmdlamin93 1 year ago
Cool!
iviewthetube 5 years ago