Stack Blocks Past the Table's Edge

Stack identical blocks so each overhangs the one below by as much as balance allows: 1/2, then 1/4, then 1/6 of a block. The overhangs shrink toward zero, yet their sum, half the harmonic series, grows without limit: with enough blocks the top one can stick out as far as you like.

How it works

The divergence test and the harmonic series

In one sentence

If $\lim a_n \ne 0$ , then $\sum a_n$ diverges (divergence test). The converse fails: the harmonic series $\sum\frac1n$ has $a_n \to 0$ but diverges, since $\sum_{n=1}^N\frac1n \ge \int_1^{N+1}\frac{dx}{x} = \ln(N+1)$ .

General formula

{\htmlData{sym=0}{d_n}} = \frac{1}{2}\left(1 + \frac12 + \frac13 + \cdots + \frac1{\htmlData{sym=2}{n}}\right) = \frac{{\htmlData{sym=1}{H_n}}}{2} \;\to\; \infty \quad\text{even though}\quad \frac{1}{2{\htmlData{sym=3}{k}}} \to 0

Terms shrinking is necessary, not enough. The harmonic terms shrink so slowly that you can always group enough of them to add another half.

What the symbols mean

$d_n$
Overhang
How far the top block sticks out past the table, in block lengths.
$H_n$
Harmonic number
1 + 1/2 + … + 1/n.
$n$
Blocks
How many blocks are stacked.
$k$
Block position
Counting from the top.

For the stack to balance, the blocks above each block must have their center of mass over its edge. That allows block $k$ (from the top) to overhang the next by $\frac{1}{2k}$ .
The total overhang of $n$ blocks is $d_n = \frac12H_n$ , half the $n$ -th harmonic number.
The terms $\frac{1}{2k}$ go to 0, so the divergence test says nothing: terms going to zero doesn't make a series converge.
Comparing with $\int\frac{dx}{x} = \ln x$ shows $H_n \approx \ln n$ : it grows forever, just very slowly. Four blocks clear one block length; 31 clear two; 227 clear three.

Learn the math

The Divergence TestCalc II · Lesson 27, Sequences and Series The Integral TestCalc II · Lesson 29, Sequences and Series