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Today, while solving a question in my homework, I stumbled upon a strange thing that I couldn't wrap my head around. I am not more than an amateur in mathematics, so I could barely understand the few other articles on SE with the same pattern of the question.

Consider, $$S = 1 + 9 + 9^2 + 9^3 + 9^4 + \dots \tag{1} $$ $$9S = 9 + 9^2 + 9^3 + 9^4 + \dots \tag{2} $$ Subtracting we get , $-8S = 1$

Or , $S = -\frac{1}{8}$

but clearly, $S = 1 + 9 + 9^2 + ...> 0$. So how can $S$ be $-\frac{1}{8}$?

I asked my teacher about this, who said that this series gives "accurate" results only when $|x| < 1$. However, I am yet to understand what actually went wrong in my derivation. Clearly $S$ tends to infinity, but then one line below, due to some atrocious mathematics done by me, it says otherwise.

I want to know what the "atrocious mathematics" is in this case. Almost seems like the proofs concluding $0 = 1$, where they commit a fallacy by dividing by $0$.

M.B.
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    In general it is not true that $x(y_1+y_2+y_3+\cdots)=xy_1 + xy_2 + xy_3 + \cdots$. The series is the limit of the partial sums so you have to be sure the partial sums make sense before you distribute. – CyclotomicField Oct 29 '24 at 20:29

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Since you've just started learning about series, you've probably just seen formulas like $1 + x + x^2 + \cdots + x^n + \cdots = \frac{1}{1-x}$ without any context whatsoever.

The fallacy that you're making is assuming that $S$ even exists in the first place! Not all series actually have a finite sum:

  • Series with a sum are called convergent
  • Series without a sum are called divergent

The series $S=1 + 9 + 9^2 + \cdots$ is in fact divergent. So multiplying $S$ by nine makes no more sense than asking "What is the color red times 9?"

When your teacher says that "this series gives 'accurate' results only when $|x|<1$" what they really should be saying is that the series only has a sum when $|x|<1$.

mathperson314
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  • Worth pointing out in OP's case that they have $x=9$, which is far away from the convergent range. – Randall Oct 29 '24 at 20:24
  • You could use the extended real numbers (that is, you include $\infty$ and $-\infty$) so that $S$ is well-defined and equals $\infty$. But then you can't subtract the two equations, as $\infty - \infty$ is not defined (it's like how you can't just divide by a variable when working with real numbers, because the variable might be zero). – MartianInvader Oct 29 '24 at 20:54
  • Thanks for your answer . So the main takeaway is that when writing infinite series like this , we should always specify $|x|<1$ , like we say $x \neq 0$ when dividing by 0 ? Otherwise we could conclude that the sum doesn't even exist . – M.B. Oct 29 '24 at 21:19
  • That is correct, @MikeBillings. The analogy is apt. In both cases you're trying to play with mathematical objects that don't exist. – mathperson314 Oct 29 '24 at 21:21