2
$\begingroup$

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:

enter image description here

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

$\endgroup$
3
  • $\begingroup$ Each factor is a geometric progression. $\endgroup$
    – saulspatz
    Jun 14 at 21:40
  • 1
    $\begingroup$ $\displaystyle (1 + r + r^2 + \cdots + r^n)(1-r) = \left(1 - r^{n+1}\right).$ Therefore, $\displaystyle (1 + r + r^2 + \cdots + r^n) = \frac{1 - r^{n+1}}{1-r}.$ $\endgroup$ Jun 14 at 21:54
  • $\begingroup$ 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. $\endgroup$ Jun 14 at 22:21
0
$\begingroup$

$\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 $|z| < 1$.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $z^m$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.