# Help understanding sequence proof

Prove $$(1 + z + z^{2} + \ldots + z^{9}) \cdot (1 + z^{10} + z^{20} + \ldots + z^{90}) \cdot (1 + z^{100} + z^{200} + \ldots + z^{900}) \cdot \ldots = \frac{1}{1 - z}$$.

For $$|z|<1$$.

De Souza and Silva give the following proof in the book Berkeley Problems in Mathematics:

But I don't understand how they got there. It seems to me several steps are missing.

• Each factor is a geometric progression. Jun 14 at 21:40
• $\displaystyle (1 + r + r^2 + \cdots + r^n)(1-r) = \left(1 - r^{n+1}\right).$ Therefore, $\displaystyle (1 + r + r^2 + \cdots + r^n) = \frac{1 - r^{n+1}}{1-r}.$ Jun 14 at 21:54
• Note, if you interpret this in terms of generating functions this is equivalent to saying that each positive integer has a unique base-10 expansion. Jun 14 at 22:21

$$\require {cancel}$$

You can verify for yourself that for any $$w \in \mathbb C$$ then $$(1-w)(1 + w + w^2 + ..... + w^{n-1}) = 1-w^n$$.

So if $$w \ne 1$$ then $$\frac {1-w^n}{1-w}= (1 + w+w^2 + ... + w^{n-1})$$ and if $$w = z^{k}$$ then $$(1 + z^k + z^{2k} + .... + z^{mk}) = \frac {1 - z^{(m+1)k}}{1-z^k}$$

And so every $$1 + z^{10^k} + z^{2\times 10^k} + .... + z^{9\times 10^k}=\frac {1-z^{10^{k+1}}}{1- z^{10^k}}$$

This is a very well known result and the text assumes the reader should have seen that result (or something very similar to it) many many times before.

So

$$(1 + z + z^{2} + \ldots + z^{9}) \cdot (1 + z^{10} + z^{20} + \ldots + z^{90}) \cdot (1 + z^{100} + z^{200} + \ldots + z^{900}) \cdot \ldots (1 + z^{10^k} + z^{2\times 10^k} + .... + z^{9\times 10^k})=$$

$$\frac {1 - z^{10}}{1 - z}\cdot \frac {1- z^{100}}{1-z^{10}} \cdot \frac {1-z^{100}}{1-z^{100}} \cdot .... \cdot\frac{1 + z^{10^{k+1}}}{1-z^{10^k}}=$$

$$\frac {\cancel{1 - z^{10}}}{1 - z}\cdot \frac {\cancel{1- z^{100}}}{\cancel{1-z^{10}}} \cdot \frac {\cancel{1-z^{100}}}{\cancel{1-z^{100}}} \cdot .... \cdot\frac{1 + z^{10^{k+1}}}{\cancel{1-z^{10^k}}}=$$

$$\frac {1+z^{k+1}}{1-z}$$

Now if (and only if) $$|z| < 1$$ then $$\lim_{k\to \infty} z^{k+1} = 0$$ and $$\lim_{k\to\infty} \frac {1 - z^{10^{k+1}}}{1-z} =\frac 1{1-z}$$.

And so if $$|z| < 1$$ then

$$\lim_{k\to \infty} (1 + z + z^{2} + \ldots + z^{9}) \cdot (1 + z^{10} + z^{20} + \ldots + z^{90}) \cdot (1 + z^{100} + z^{200} + \ldots + z^{900}) \cdot \ldots (1 + z^{10^k} + z^{2\times 10^k} + .... + z^{9\times 10^k})=$$

$$\lim_{k\to \infty} \frac {1-z^{10^{k+1}}}{1-z} = \frac 1{1-z}$$

And as this does converge we may so

$$(1 + z + z^{2} + \ldots + z^{9}) \cdot (1 + z^{10} + z^{20} + \ldots + z^{90}) \cdot (1 + z^{100} + z^{200} + \ldots + z^{900}) \cdot \ldots (1 + z^{10^k} + z^{2\times 10^k} + .... =$$ (which is an infinite product)

$$\lim_{k\to \infty} (1 + z + z^{2} + \ldots + z^{9}) \cdot (1 + z^{10} + z^{20} + \ldots + z^{90}) \cdot (1 + z^{100} + z^{200} + \ldots + z^{900}) \cdot \ldots (1 + z^{10^k} + z^{2\times 10^k} + .... + z^{9\times 10^k})$$

which we just saw was $$\frac 1{1-z}$$.

But this requires that $$|z| < 1$$.

Note that $$\prod_{i=0}^{n-1}\left(\sum_{j=0}^9z^{10^ij}\right)=\prod_{i=0}^{n-1}\left(\frac{z^{10^{i+1}}-1}{z^{10^i}-1}\right)=\frac{z^{10^n}-1}{z-1},$$ wich goes to $$-1/(z-1)=1/(1-z)$$ as $$n\to \infty$$.

In the first step we've used that $$(z^{10^i}-1)\left(1+z^{10^i}+\cdots+z^{9\cdot 10^i}\right) = z^{10^{i+1}}-1,$$ whence $$\sum_{j=0}^9z^{10^i\cdot j}=\frac{z^{10^{i+1}}}{z^{10^i}-1}$$, and in the second step, we've used that $$\frac{z^{10}-1}{z-1}\cdot \frac{z^{100}-1}{z^{10}-1}\cdots \frac{z^{10^n}-1}{z^{10^{n-1}}-1} = \frac{z^{10^n}-1}{z-1},$$ because most factors cancel.

Alternatively, any integer $$m$$ has a unique representation in base 10: $$m=a_0+a_1\cdot 10+a_2\cdot 10^2+\cdots$$. For example $$103=3+\cdot 10^0+1\cdot 10^2$$, which in exponents would look like $$z^{103}=z^{3\cdot 1}\cdot z^{0\cdot 10}\cdot z^{1\cdot 100}$$

Convince yourself then that every integer power of $$z$$ occurs exactly once the infinite product, once expanded. You can formally prove this by expanding a finite product that has sufficiently many terms for a given $$z^m$$.