# Evaluating $\frac 9 {10}\cdot\frac {99} {100}\cdot\frac {999} {1000}\cdots$ [duplicate]

$\displaystyle\frac 9 {10}\cdot\frac {99} {100}\cdot\frac {999} {1000}\cdots=?$

Usually, product of infinite many numbers which are less than 1, is 0. But How about this time?

Thank you.

• It does not appear that the limit is zero. WolframAlpha gives a very interesting output. – preferred_anon Apr 13 '13 at 13:12
• ya,I guess $\ge 0.89$ – chenbai Apr 13 '13 at 13:41

It's $\displaystyle \left(\frac{1}{10};\frac{1}{10}\right)_\infty=\sqrt[24]{10}\eta\left(-\frac{\log 10}{2\pi i}\right)\approx \href{http://www.wolframalpha.com/input/?i=%2810%5E1%2F24%29*DedekindEta%5B-+Log%5B10%5D+%2F+%282*Pi*I%29%5D}{0.890010099998999000000100009999999989999900}...$

See q-Pochhammer symbol and Dedekind eta function for details.

Proving that the product converges to a positive value can be proven elementarily. We have

$$\prod_{k=2}^n\left(1-\frac{1}{k^2}\right)=\left[\frac{\bf 2-1}{\color{Blue}{2}}\frac{\color{Blue}{3-1}}{\color{Red}{3}}\frac{\color{Red}{4-1}}{\color{DarkGreen}{4}}\cdots\frac{\color{DarkOrange}{n-1}}{\bf n}\right]\left[\frac{\color{Blue}{2+1}}{\bf 2}\frac{\color{Red}{3+1}}{\color{Blue}{3}}\frac{\color{DarkGreen}{4+1}}{\color{Red}{4}}\cdots\frac{\bf n+1}{\color{DarkOrange}{n}}\right]$$

$$=\frac{2-1}{n}\frac{n+1}{2}\longrightarrow \frac{1}{2}\quad\textrm{as}~n\to\infty.$$

By comparison, $k^2<10^{k-1}\implies 1-1/k^2<1-10^{-k}$, which shows our value converges to $>1/2$.

• Ah, those colors :-) – Cortizol Apr 13 '13 at 13:38
• Very colourful answer. – Ambesh Apr 13 '13 at 14:28

The product of of a series infinitely many numbers less than 1 may be 0, but your series is not one of infinitely many numbers less than one. The last number in your series is 1.

Your series $$\displaystyle\frac 9 {10}\cdot\frac {99} {100}\cdot\frac {999} {1000}\cdots$$

is equivalent to:

$$\displaystyle 0.9\times0.99\times0.999\times\cdots\times0.\overline9$$

the final term of which, $0.\overline9 = 1$. Since the numbers in the series increase towards one, this series appears to converge toward a non-zero value. Multiplication of the first few terms appears to confirm this: $$0.9\times0.99 = 0.891$$ $$0.891\times0.999=0.890109$$ $$0.890109\times0.9999=0.8900199891$$ $$0.8900199891\times0.99999=0.890011088900109$$ $$0.890011088900109\times0.999999=0.8900109889020099891$$ $$0.8900109889020099891\times0.9999999=0.8900101098880002109889900109$$

• There is no "last term" in an infinite series or an infinite product - these things are defined with limits. Every single factor in the product is strictly less than $1$. That that the factors increase and converge to $1$ is insufficient to conclude anything. For example, $\prod(1-1/p_k)$, where $p_k$ is the $k$th prime number, also has terms increasing towards $1$, but the partial products converge to $0$ (albeit slowly - the asymptotic rate of decay is the subject of Merten's 3rd theorem). – anon Apr 13 '13 at 13:34
• A more elementary counterexample to this line of reasoning: $e^{-1/k}$ is increasing as a function of $n$, and converges to $1$ in the limit, but $\prod_{k=1}^n e^{-1/k}=\exp\big(-(1+1/2+\cdots+1/n)\big)$ converges to $0$ since the harmonic series diverges. In fact, since $H_n=\log n+\gamma+O(1/n)$, the partial products are roughly $e^{-\gamma}/n$. – anon Apr 13 '13 at 13:52