# 1 + 10 + 100 + 1000 + = -1/9

## Main Question or Discussion Point

$$S = 1 + 10 + 100 + 1000 + 10000 + ...$$

$$10S = 10 + 100 + 1000 + 10000 + 100000 + ...$$

$$S - 10S = (1 + 10 + 100 + 1000 + 10000 + ...) - (10 + 100 + 1000 + 10000 + ...)$$

$$-9S = 1 + (10 - 10) + (100 - 100) + (1000 - 1000) + (10000 - 10000) ...$$

$$-9S = 1 + 0 + 0 + 0 + 0 + 0 ...$$

$$-9S = 1$$

$$S = -1/9$$

What's wrong (or right) with this?

Thanks,
Unit

Hurkyl
Staff Emeritus
Gold Member
Also, I believe that sum converges as an ordinary infinite sum in the 2-adics and the 5-adics.

(And, of course, it does not converge as an ordinary infinite sum in the reals!)

CRGreathouse
Homework Helper
I would have [intuitively] expected it to converge in all the p-adics. Am I wrong?

Hurkyl
Staff Emeritus
Gold Member
In any other p-adic field, the terms don't converge to zero!

CRGreathouse
Homework Helper
In any other p-adic field, the terms don't converge to zero!

Char. Limit
Gold Member
I'm pretty sure that you can't pair up terms in an infinite sum.

Redbelly98
Staff Emeritus
Homework Helper
$$S = 1 + 10 + 100 + 1000 + 10000 + ...$$
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What's wrong (or right) with this?

Thanks,
Unit
I'm pretty sure that you can't pair up terms in an infinite sum.
The way I remember it, proofs like this actually say something like:
If S exists, then S = 1 + 10 + 100 + ...​
So if S does not exist, then the remaining statements do not necessarily hold true.

If S exists, then S = 1 + 10 + 100 + ...​
So if S does not exist, then the remaining statements do not necessarily hold true.
Brilliant! I had completely forgotten about variables and their related hypothetical syllogisms. Thanks!