Showing that 0.999... = 1 using geometric sums

If you have browsed a forum where math is discussed, it is likely you have seen people claiming both that \(0.999... = 1\) and \(0.999... \neq 1\). Today I will show you that the first one is correct using geometric sums. You don’t need to take my word for it, you will be able to prove it yourself.

Geometric Sums

“What is a geometric sum?”, you may ask. Keep calm, I will get you covered. A geometric sum is a sum where the first term is a number, let’s call it \(a_1\) and the second term is a constant, \(k\), times the first term, so \(a_1*k\). The third term is the second term times the same constant so it becomes \(a_1*k*k = a_1*k^2\). And then it continues like that for as long as we want to calculate the sum. If we want to calculate the sum of the first 5 terms the last one would be \(a_1*k^4\) because the first term don’t have a \(k\) in it. Generally the last term thus becomes \(a_1*k^{n-1}\) if we want to take the sum of the \(n\) first terms. A geometric sum is of the form:

\[S_n = a_1 + a_1*k + a_1*k^2 + a_1*k^3 + a_1*k^4 + ... + a_1*k^{n-2} + a_1*k^{n-1}\]

Here \(S_n\) stands for “the sum of the \(n\) first terms”. The terms are quite similar so it isn’t to farfetched to think that there is a formula to calculate the the sum given only \(a_1\), \(k\) and \(n\). There is in fact a formula but I won’t just give it to you, we will derive it ourself to have a stable foundation for our argument.

Deriving the formula for a Geometric Sum

We start out with the equation for a general geometric sum:

\[S_n = a_1 + a_1*k + a_1*k^2 + a_1*k^3 + a_1*k^4 + ... + a_1*k^{n-2} + a_1*k^{n-1}\]

Then we multiply both sides by \(k\):

\[k*S_n = a_1*k + a_1*k^2 + a_1*k^3 + a_1*k^4 + a_1*k^5 + ... + a_1*k^{n-1} + a_1*k^n\]

The reason we did this is because now we can see that the right side of the equation is nearly the same as it was before except \(a_1\) is missing and we have a new term, \(a_1*k^n\). We can then utilize that fact to rewrite the equation like this:

\[k*S_n = S_n + a_1*k^n - a_1\]

This is much neater, right? Now all we have to do is to solve for \(S_n\). We start by subtracting \(S_n\) from both sides:

\[k*S_n - S_n = a_1*k^n - a_1\]

Then we factor both sides by them self:

\[S_n(k - 1) = a_1(k^n - 1)\]

Dividing by \((k-1)\) will give us our formula:

\[S_n = a_1 \frac{k^n - 1}{k - 1}\]

There we have it, a simple formula for calculating a geometric sum.

Let’s tackle the question

I said we would show that \(0.999... = 1\) so let’s cut to the action. First what does \(0.999...\) really mean? In this context it means that the pattern shown will continue forever. In other words that there is an infinite amount of 9’s after the decimal sign. Just a quick reflection here: Is there any real number between 1 and 0.999…? Independent of your answer, what implications would that have? Think about it, and experiment. That’s how new mathematics are invented, people playing around with the rules of math. “What happens if I ignore that?” and “How would this work if it was like this” are questions reasonable to wonder about. Complex numbers (they are as real as the real numbers, not imaginary in any way) were found this way, when someone thought “What if negative numbers have a square root” and just rolled along.

Must have been an uneven cut because we seems to have sidetracked a bit from the action. 0.999… can if we separate every digit into a 9 times a power of a 10th be written as:

\[0 + 9 \frac{1}{10} + 9 \frac{1}{10^2} + 9 \frac{1}{10^3} ...\]

This looks looks like a geometric sum, who could have guessed? \(a_1\) in this case is \(9\frac{1}{10}\), \(k\) is \(\frac{1}{10}\) and \(n\) is \(\infty\). We have a slight problem here: \(n\) is \(\infty\). Let’s see how our formula handles that.

\[S_n = a_1 \frac{k^n - 1}{k - 1}\]

If \(k\) is larger than or equal to 1, then the sum diverges and it doesn’t have a defined sum. If on the other hand \(k < 1\), then the sum converges. We can in fact simplify the formula when both \(n = \infty\) and \(k < 1\) because \(k^n\) then becomes 0 at the limit:

\[lim_{n\to\infty} a_1\frac{k^n-1}{k-1} = a_1\frac{0-1}{k-1} = \frac{-a_1}{k-1} = \frac{-a_1}{-(1-k)} = \frac{a_1}{1-k}\]

There we have a really nice and simple formula. It is quite remarkable that we can calculate something that has an infinite component using such a simple and beautiful formula. If we now insert our values into it we get:

\[\frac{a_1}{1-k} = \frac{\frac{9}{10}}{1-\frac{1}{10}} = \frac{\frac{9}{10}}{\frac{9}{10}} = 1\]

There we have it! We have showed that \(0.999... = 1\) using a geometric sum. If someone ever tells you that it isn’t true, prove it to them yourself. You have all the tools you need for that now.

Can you come up with any other “debate-able” sequence of decimals that can be proven to be something using geometric sums? If you do, please let me know in the comments. Have a nice day and remember that the human fantasy is limitless (but not undefined).

Written on July 31, 2018