$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{\infty}=?$

Question

One of my friends has given me the following problem:

$$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{\infty} = ?$$

He said that the answer is $1$. He has given his argument as that this is a part of the whole, and if we add together all the part, then we'd have a $1$. But said, disagreed with him that it'd not be $1$, rather it'd be $\frac{\infty - 1}{\infty}$. I'm not sure if this is correct. But I have given the argument with this view:

1. If we add together only the first two fractions, that is $\frac{1}{2}$ and $\frac{1}{4}$, then the sum will be $\frac{3}{4}$, which is equal to $\frac{4-1}{4}$.

2. Likewise, if we add together the first three, four, five, then the sum will be $\frac{x-1}{x}$, where $x$ is the denominator of last fraction. These happen because every time divide ramainder(whole in the first case) in the equal halves, there remain one part of the equal halves and after we add together all the part(follows the pattern, never break the pattern) there remain $\frac{1}{x}$ of the whole part. So we can nonetheless say this that this only became 1(the whole), if we add add the remainder in the sum, that's mean there will be $2$ of the last fraction. But that will break the pattern. But he argued with me and wanted me to ask the question here. I also said that if we round it then it will be $1$, rather it never will be $1$.

So, what is your opinion about it? If included both, argument and mathematical calculation, then that will be great.

Answer 1

There is no such thing as $$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{\infty}.$$ On the other hand, there is something commonly notated $$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$$ or, equivalently, $$\sum_{n=1}^\infty \frac{1}{2^n}. $$

By definition these two latter notations mean the limit of $$ \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{2^k} $$ as $k$ "goes to infinity", or grows without bound. The "partial sum" up to $\frac{1}{2^k} $ equals $$ \frac{2^k-1}{2^k} = 1-\frac{1}{2^k} $$ and the limit of this is exactly $1$, because we can make $1-\frac{1}{2^k}$ be as close to $1$ as we want by taking $k$ large enough. This is the definition of what a limit means.

You seem to think that in order to take the limit one can simply plug the $\infty$ symbol into the formula instead of $k$. But that is nonsense; $\infty$ is not a number -- and even if you take a limit at an actual number rather than infinity, simply plugging that number into the formula is not guaranteed to yield the limit.

Answer 2

The only way of making sense of the sum you propose is to interpret $\infty$ as an element in a suitable number system, as opposed to its generic use in various fields (such as $\lim_{x\to\infty}$ in calculus). In such a number system the sum you propose indeed has a well-defined meaning (called a hyperfinite sum) and it does fall infinitesimally short of $1$, as you suggested.

Note that it would be more correct to list the last term in your sum as $\frac{1}{2^{\infty}}$.