Can you please explain why 1/2 + 1/4 + 1/8 + ... = 1?

Question

Thanks in advance!

Answer 1

Sure!

Let x = 1/2 + 1/4 + 1/8 + ...

We have

x = 1/2 + (1/2)(1/2 + 1/4 + 1/8 + ...)
x = 1/2 + (1/2)x

This reduces to

x - (1/2)x = 1/2

THUS

(1/2)x = 1/2 or x = 1!

The attack shown here is very similar to my answer to John J's question about 0.999... = 1.

Have a good day

Answer 2

Another way to look at it (a variation of the above answer)
x = 1/2+1/4+1/8+...
2x=1+1/2+1/4+1/8+...
2x-x=(1+1/2+1/4+1/8+...)-(1/2+1/4+1/8+...)
x=1

Answer 3

Assume
F(x) = 1/2+1/4+1/8+1/16+...............1/2^(K) [k->infinity]
Then (1/2)F(x) = 1/2 [ 1/2+1/4+1/8+1/16...........+1/2^(k)+1/2^(infinity+1)]
(1/2)F(x) = 1/4 +1/8 +1/16 +1/32.............+1/2^(k)+ 1/2^(1+k)

Then
F(X) - (1/2)F(x) =(1/2)F(X)
=1/2 +1/4 +1/8 + 1/16...............+1/2^(k)
- 1/4 -1/8 - 1/16 - 1/ 32-...-1/2^(k)-1/2^(k+1)
=1/2- 1/2^(k+1)
because K->infinity, 1/2^(k+1)=0
=1/2

because
(1/2)F(x) =1/2
F(x) = (1/2) / (1/2) = 1
#

Answer 4

This is a geometric series, the first term is 1/2 and the commonn quocient is 1/2.

We now that, if |q| <1, then the series a + aq + aq^2 + aq^3.....converges to a/(1-q). In your case, the series converges to (1/2)/(1 -1/2) =(1/2)/(1/2) = 1.

Answer 5

I will give you another example to ponder. The last digit of
π in binary.
Since binary representations have only two values 1 and 0, then the last digit of π in binary must be either a 1 or 0.
If π is a single value constant than the last digit of π must be 0 and it will end with a infinite number of zero's.

Because if it ended with 1 we can always tack on an infinite number of 0's after without changing the value of π. This isn't true with any combination of 1's and 0's.

Otherwise if it end with any arbitrary combination of 1's and 0's then it isn't a single-value constant, but an arbitrarily small region of multiple indistinguishable values and isn't a single value constant.

Answer 6

Most of the other people have answered this correctly, so I will not respond to the actual question, but I did want to post to tell people to ignore the last part of Larry C's answer.

Infinity over infinity is *absolutely by no means certainly* NOT one. It can be, but its the same as saying x can equal 1. x can also equal many many other things.

Answer 7

Draw an empty box (a square).
Now fill in 1/2 of the box.
Now fill in 1/2 of the empty space, which is 1/4 of the box.
You now have filled in 1/2 + 1/4 of the the box, or 3/4.

Now fill in another 1/2 of the empty space.
You now have filled in 1/2 + 1/4 + 1/8 of the the box, or 7/8.

You can see that as you approach infinity, the box will be filled in.

Or, the numerator is infinity and the denominator is infinity.
Infinity over infinity is one.

Answer 8

the REAL question is does .999999 = 1?

see this web page!

Answer 9

Logically this equation will never equal 1 but it will continue to approach it, ad infinitum....

Answer 10

because it's not 1, it's 7/8.........