$S=1+10+100+100+10000+... = -1/9$? How is that

Solution 1:

This does not hold for real numbers, the numbers you are probably used to working with. Rather it works with a different type of number, called a p-adic number. I will get to what these are in a moment.

But first, what is a number? One might be tempted to give examples, such as $1$ or $-3/4$. But these are just symbols, which have no meaning on their own. Rather, mathematicians have defined them in a way that they all agree upon, which allows them to be added, multiplied, etc. It is by defining them that we give them meaning.

But why do we define them the way we do? We could just as easily define them in other ways; it just so happens that the typical definition is often the most useful. But this is not always the case. The p-adic numbers are a different way of defining "number".

To explain, I'll take the special case of 10-adic integers, which is where the equation you are asking about applies. The normal integers are what we are used to using daily, and if we take two integers $a$ and $b$ we can talk about how "close" $a$ is to $b$. Specifically, we say that $a$ is "close" to $b$ if they differ by a small power of ten, i.e. $40$ is sort of close to $30$ because $40-30=10=10^1$ while $4$ is very close to $3$ because $4-3=1=10^0$ (the same process works for numbers which do not differ by an integer power of ten if we allow non-integer powers). But why do we define "closeness" this way? What if we said $a$ is close to $b$ if they differ by a large power of ten? For example, $40,000$ would be sort of close to $30,000$ because $40,000-30,000=10,000=10^4$, while $4,000,000$ would be very close to $3,000,000$ because $4,000,000-3,000,000=1,000,000=10^6$. If we define "closeness" in this way, we get the 10-adic integers.

Now for the series you posted. Multiplying on both sides by $9$ gives $9+90+900+9000+90000+… = -1$. Well, we can see that this is giving us $\cdots999999 = -1$, with an infinite number of $9$'s to the left. But $9-(-1)=10=10^1$, $99-(-1)=100=10^2$, etc. so as we add more $9$'s onto the left we are getting closer and closer to $-1$, using the way we defined "closeness" for the 10-adic integers. In fact, we get infinitely close to $-1$, so this series must in fact be equal to $-1$ in the 10-adic integers. While this is not a rigorous argument, I hope it provides you with a rough sense of what's going.

Edit: Tying in what Ross said in his answer, the reason the formula $1/(1-10) = 1 + 10 + \cdots$ works in the 10-adic integers is that each successive term is getting closer to zero using the 10-adic notion of "closeness", so we can in fact ignore sufficiently late terms.

Solution 2:

There are many formulas that fall in this category. The sum is not convergent, but you can do a formal manipulation that seems to sum it. Maybe the simplest is 1-1+1-1+1-1 ... If you represent it as $\sum (-1)^i$ you can sum it as $\frac {1}{1-(-1)}=\frac{1}{2}$. This is exactly in the spirit of your original comment "Using $\frac{1}{1−x}$ I can put $x=10$ but that is not fair,isn't it?" No, it's not fair, as the series is not convergent in the usual real number system, and as the p-adic numbers are not "normal" you should alert your readers when you use them. The formal power series manipulation that leads to the sum of the geometric series cancels all the intermediate terms just fine, but the last term doesn't go to zero so you can't ignore it.