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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.

Jam
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Curious learner
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2 Answers2

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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.

hmakholm left over Monica
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    I think this is the only answer to actually address the asker's confusion and not just explain what a geometric series is… – Jam May 22 '18 at 10:20
  • Your ans. have appeased my confusion much. But I got stuck at a point. You said that the sum is equal $\frac{2^k - 1}{2^k}$. that's mean this is not equal to $1$. but you also said that the limit is exactly $1$.if I enlarge the $k$ then the sum will relatively very close to $1$. So please say clearly, the sum is $1$ or not. Anyway, is there a huge problem if I use $\infty$ rather then $k$. – Curious learner May 23 '18 at 03:31
  • @Curiouslearner: The partial sums, that is, the sum of only the first $k$ terms, are $\frac{2^k-1}{2^k}$, which is not equal to $1$. The limit of those partial sums is $1$ exactly. The sum of the infinite series of terms is -- by definition! -- just a shorter way to speak of the limit of the partial sums, which is $1$ no matter what we call it. – hmakholm left over Monica May 23 '18 at 03:36
  • If you use $\infty$ instead of $k$ you will at best confuse readers, because the $\infty$ symbol is not generally used as a variable that takes finite values, which is what $k$ is here. – hmakholm left over Monica May 23 '18 at 03:37
  • Ohh, your word is confusing me a little and going over my head. You said that the sum of $\frac{2^k-1}{2^k}$ is not equal to $1$. I understood till this. After this, I can't catch you what the sum will be as you if $k$ go on and on, that's mean $k=\infty$. so please say what the sum will be if $k$ goes without bound. Plz don't confuse me again!! – Curious learner May 23 '18 at 14:39
  • @Curiouslearner: No, there is nothing here that means $k=\infty$. The value of $k$ is always an ordinary finite natural number. When we take the limit we're considering what happens when $k$ gets larger and larger, but still all we're considering there is large finite values of $k$. What happens when $k$ gets larger is that $\frac{2^k-1}{2^k}$ gets closer to $1$, and we can make it as close to $1$ as we want by making $k$ sufficiently large (but still finite). This, by definition, is what it means for the limit to be $1$. There is no $\infty$ anywhere in this computation. – hmakholm left over Monica May 23 '18 at 15:02
  • (... except that the symbol $\infty$ occurs in the notation $\sum\limits_{k=1}^\infty$, but that is purely a suggestive way to say "and there is no upper end of the summation" -- it does not mean that $\infty$ stands for anything that could possibly be the value of $k$. The notation could equally well have been, for example, $\sum\limits_{k=1}^{\vdots}$). – hmakholm left over Monica May 23 '18 at 15:05
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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 $\frac{1}{2^{\infty}}$.

Mikhail Katz
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