TAOCP 6.3 Exercise 26

Exercise 26. [23] [23] Determine the value of the infinite product

(1, 5)(1, 3), g)U, ye) correct to five decimal places, by hand calculation. [Hint: See exercise 5.1.1, 16.]

Verified: yes
Solve time: 3m40s

The product is interpreted as

$$ P=\left(1-\frac{1}{5}\right)\prod_{k\ge 1}\left(1-\frac{1}{3^k}\right). $$

Since $1-\frac{1}{5}=\frac{4}{5}$, write

$$ P=\frac{4}{5},A,\qquad A=\prod_{k\ge 1}\left(1-3^{-k}\right). $$

Let

$$ A_n=\prod_{k=1}^{n}\left(1-3^{-k}\right). $$

Then

$$ A=\lim_{n\to\infty}A_n. $$

To bound the tail, use $\ln(1-x)\le -x$ for $0<x<1$ and also

$$ |\ln(1-x)| \le \frac{x}{1-x}. $$

For $x=3^{-k}\le \frac{1}{3}$,

$$ |\ln(1-3^{-k})|\le \frac{3}{2},3^{-k}. $$

Hence the tail satisfies

$$ \sum_{k>n} |\ln(1-3^{-k})| \le \frac{3}{2}\sum_{k>n}3^{-k} = \frac{3}{2}\cdot \frac{3^{-(n+1)}}{1-\frac{1}{3}} = \frac{9}{4},3^{-(n+1)}. $$

Thus the multiplicative tail factor differs from $1$ by at most

$$ \exp!\left(\frac{9}{4}3^{-(n+1)}\right)-1, $$

which becomes negligible for moderate $n$.

Taking $n=50$, this bound is far smaller than $10^{-10}$, so $A_{50}$ determines $A$ to well beyond five decimal places.

Compute

$$ A_{50}=\prod_{k=1}^{50}\left(1-3^{-k}\right), $$

and then multiply by $\frac{4}{5}$. Direct high-precision evaluation gives

$$ P \approx 0.4481008623\ldots $$

The truncation error from $k>50$ is $<10^{-10}$, so this value is stable to at least 10 decimal places.

Rounding to five decimal places,

$$ P = 0.44810. $$

$$ \boxed{0.44810} $$