Is there a closed form for 0.9 * 0.99 * 0.999 * ...?

Asked Jul 06 '18 at 18:13

Active Jul 06 '18 at 18:15

Viewed 328 times

I'm trying to find a pretty closed form for:

$$0.9 \cdot 0.99\cdot 0.999 \cdots=\prod_{n=1}^\infty \frac{10^n-1}{10^n}$$

The Nth partial product can be expressed as:

$$\frac{(10-1)(10^2-1)\cdots(10^N-1)}{10^{\frac{n(n+1)}{2}}}$$

But it seems impossible to generalize the expansion of the numerator polynomial in 10.

Wolfram Alpha computes the following approximation:

$≈0.89001009999899900000010000999999998999990000000000100000099999999999989999999000000000000001000000009999999999999999899999999900000000$

Hence I strongly believe that a closed form exists.

edited Jul 06 '18 at 18:15

Stefan4024

asked Jul 06 '18 at 18:13

greenturtle3141

10

It is $\phi(1/10)$ where $\phi(q)$ is Euler function. See Evaluate $\prod_{n=1}^\infty \frac{2^n-1}{2^n}$ – Sil Jul 06 '18 at 18:21
7

The regularity you see in the decimal expansion is related to the Euler identity, also known as the pentagonal number theorem: $$\phi(q)=\sum_{n=-\infty}^\infty (-1)^n q^{(3n^2-n)/2};.$$ – joriki Jul 06 '18 at 18:24

0 Answers0