Does $\frac12+\frac14+\frac18+\dots$ equal $1$? Suppose I have something with length one unit. I divide it to two equal length $0.5$ unit and put left one in my left side and right one in my right side. I then do same for my right side and contact result's left one to my left side and replace my right side with it's right one.
Even if I do these infinitely, there's something at my right side remaining. So it seems that $$\frac12+\frac14+\frac18+\dots$$ never taste value $1$. Right?
NOT DUPLICATE: Actually answers at here does not answer my question.
because if I accept them then I have
$$\frac12+\frac14+\frac18+\dots=\frac23+\frac29+\frac2{27}+\dots=1$$
Then I can define
$$A_n = \frac12+\frac14+\frac18+\dots+\frac1{2^n}$$
And
$$B_n = \frac23+\frac29+\frac2{27}+\dots+\frac2{3^n}$$
Then, it's quite can be seen that:
$$A_n<B_n , n\in\mathbb N$$
Then how could I accept that both series can reach each other at value $1$?!
 A: It is true, that $$\sum_{k=1}^\infty \frac{1}{2^k} = 1$$
But only, if you take the whole, infinite series.
If you only take finitely many terms, let's say you have $\sum_{k=1}^n\frac{1}{2^k}$. This never equals $1$, but $1-\frac{1}{2^{n+1}}$.
So with only finitely many terms, you'll never reach $1$.
A: I think the trouble here is that "equal" is a very strong word for what we're talking about. It's a question of being specific with definitions. The series you've effectively written down:
$$
\sum_{n=1}^\infty \frac{1}{2^n}
$$
Is said to converge to 1, but not usually said to equal 1. This name much more accurately captures what you spoke of, that you never quite reach 1 after any finite number of "halvings", but that you clearly are heading that way.
This all comes from the fact that normally the definition of such an infinite series is the limit of the related sequence, so:
$$
\sum_{n=1}^\infty \frac{1}{2^n} = \lim_{m \rightarrow \infty} \sum_{n=1}^m \frac{1}{2^n}
$$
A: Your "1/2+1/4+1/8+..." is by definition the limit of the sequence
$$S_n={1\over 2^1}+{1\over 2^2}+\cdots+{1\over 2^n}={2^n-1\over 2^n}$$
So, each $S_n<1$, but $\lim_{n\to\infty}S_n=1$.
A: Do you have any problem with $0.\overline{3} = \dfrac{1}{3}$?  The left hand side really represents the sum $\frac{3}{10}+\frac{3}{100}+\frac{3}{1000}+ \dots$, and so the same problem exists here.  The very nature of real numbers is that you can take limits of sequences of them.  In some sense, all any real number is is a sequence of better and better approximations.  This corresponds nicely with reality:  you can never measure anything spot on, but you can continue to improve the sensitivity of you instruments to get better and better measurements.
A: I think the issue here is what we mean by 'is equal to' - does this even make sense for infinite series?  
If we take "$a+b$ is a number" (the fact that we can add two numbers together) as an assumption, then we can define finite sums with any number of terms like this:  
Assume $a_{1}+a_{2}+a_{3}+\ldots+a_{n-1}$ is a number. Then $$a_{1}+\ldots +a_{n-1}+a_{n}=(a_{1}+\ldots +a_{n-1})+a_{n}$$ is a number since this just uses our "$a+b$ is a number" assumption. This works for finite sums, but what do we do in the infinite case? Well, the natural definition is to say "What do the partial sums get close to?" - this isn't a perfectly precise definition, but it will work in this circumstance. We call this the 'limit' of the finite sums. In your case, you can see clearly that these partial sums get closer and closer to $1$, and if you follow our natural definition (and I stress again that this is not a perfect definition, but it does convey the feeling) then your infinite sum could be said to equal $1$.  
In more technical contexts, different definitions of infinite summation make more sense, but you wouldn't encounter these very often.
