# Overhanging Stacks

by Peter Jägare

In order for the stack to balance, the centre of mass of any sub-stack, consisting of all tokens from the top of the stack down to some token partway to the bottom, must be above the next token down. That is, the centre of gravity of the top seven tokens must be above the eighth token from the top, and so on. Or, in other words, if you lift off the topmost tokens from the larger stack, that smaller stack must be able to balance on its own. No surprises there, I hope.

The centre of mass of a single token is of course found in the centre of the disc. Therefore, one could place the topmost token such that its centrepoint lies on the rim of the second token from the top. Therefore, the overhang of the topmost token is equal to the radius of the token. The centre of mass of the two topmost tokens is halfway between the two tokens', so the second token from the top can be placed such that it overhangs the token below by one half of the radius. Continuing these calculations, you will find that this series continues with one third of the radius, one fourth, one fifth, and so on (I will give a proof of this below).

The series 1+1/2+1/3+1/4+… is known as the harmonic series. In the context of this puzzle, its important feature is the fact that it diverges to infinity. That is, you can reach as high a sum as you wish, by adding together a large enough number of terms (I will give a proof of this below, as well).

This, then, means that you can get as large an overhang as you wish on your stack of tokens. A metre? No problem. A kilometre? Sure, just keep stacking.

Of course, in reality there are a number of factors that makes this impossible, such as shaking human hands, lack of enough tokens, and the earth being a spheroid and not an infinite plane with a perpendicular gravitational field. Oh, and there's a ceiling in the way, too. But other than that, the size of the overhang is not inherently limited. For instance, getting an overhang that is the size of one diameter is fully possible; you can try it with coins (I suggest using at least six, though four is theoretically enough).

__Proof that the series of the overhangs is the harmonic series:__

The n^{th} token from the top will have n-1 tokens on top of it. If the mass of each token is 1, then the collective mass of those tokens is n-1. If those n-1 tokens are arranged perfectly, their centre of mass will be precisely above the rim of the n^{th} token. Or, to express this in another way, there is a mass of 1 at a distance of 0 from the centre of the n^{th} token (that's the n^{th} token itself), and there is another mass of n-1 at a distance of r from the centre of the n^{th} token (the n-1 tokens above it), where r is the radius of the tokens. Expressing it this way is useful, because we can now find out the (horizontal) distance between the centre of mass of the **n** topmost tokens and the centre of the n^{th} token. The difference between this distance and r is by how much the n^{th} token can overhang the next token down.

To find the distance between the centre of mass of the n topmost tokens and the centre of the n^{th} token, we multiply each of the two masses described earlier by the distance of that mass, add them together, and divide by the sum of the two masses:

0*1 + r*(n-1) r*n - r
------------- = -------
(n-1) + 1 n

Then, we subtract this from r to find the overhang:

r*n - r r*n r*n - r r*n - r*n + r r
r - ------- = --- - ------- = ------------- = -
n n n n n

In other words, the overhang of the n

^{th} token is r/n. So, for n = 1, 2, 3, 4, … the sequence of the overhangs is 1, 1/2, 1/3, 1/4, … which is the harmonic series.

__Proof that the harmonic series diverges to infinity:__

To show this, we will compare the harmonic series with another series which I shall now describe: The first term is 1. The second term is 1/2. The next two terms are 1/4. The next four terms are 1/8. The next eight terms are 1/16. And so on.

1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + …
1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + …

As you can see, each term of the harmonic series is either larger than or equal to the corresponding term of the other series. Therefore, the sum of the harmonic series must be at least as large as the smaller series.

Finding the sum of the smaller series is simple: We start by taking the first term, 1. Then we add the next term, 1/2. And then the next two, those two being equal to 1/2 when taken together. And then we add the next four terms, also 1/2. And the next eight, then the next sixteen, and so on, each addition being equal to 1/2. Since we can always add another 1/2, the series diverges to infinity, and therefore, so must the harmonic series.