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Is there any elementary proof that the limit of infinite product $\prod_{k=1}^{\infty}(1-\frac{1}{2^k})$ is not 0?

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Elsewhere on the internet... – The Chaz 2.0 Nov 3 '11 at 19:35
Six questions, each with at least one answer, zero accepted answers: the perfect score. – Did Nov 22 '11 at 19:13
up vote 1 down vote accepted

This is similar to the beginning of Alex's answer, however it proceeds differently.
We express the log of the product as a sum of logarithms
$$\small L=-\log(\prod_{k=1}^\infty (1-{1 \over 2^k})) = -\sum_{k=1}^\infty \log(1-{1 \over 2^k})$$ Next we expand each of the logs according to its power series and write this as a double sum; then we change order of summation and sum up columnwise; this is allowed because both directions provide convergent sums. So $$\small \begin{array} {lllllll} -\log(1-1/2)=&+1/2 & +1/2 \cdot 1/2^2 & +1/3 \cdot 1/2^3 & +1/4 \cdot 1/2^4 & \ldots \\ -\log(1-1/4)=&+1/4 & +1/2 \cdot 1/4^2 & +1/3 \cdot 1/4^3 & +1/4 \cdot 1/4^4 & \ldots \\ -\log(1-1/8)=&+1/8 & +1/2 \cdot 1/8^2 & +1/3 \cdot 1/8^3 & +1/4 \cdot 1/8^4 & \ldots \\ -\log(1-1/16)=&+1/16 & +1/2 \cdot 1/16^2 & +1/3 \cdot 1/16^3 & +1/4 \cdot 1/16^4 & \ldots \\ \cdots & & \cdots &\cdots &\cdots & \\ \hline \\ L = & 1 & + 1/2 \cdot 1/3 & + 1/3 \cdot 1/7 & + 1/4 \cdot 1/15 & \ldots \\ \hline \\ \zeta(2)=&1&+ 1/2\cdot 1/2& + 1/3 \cdot 1/3 & + 1/4 \cdot 1/4 & \dots \\ \end{array} $$ We compare L with $\small \zeta(2) $ and observe, that the absolute value each term in L is smaller than the corresponding term in $\small \zeta(2)$ thus L is a finite/converging sum $\small 0 \lt L \lt \zeta(2) \lt \infty $ and thus $\small \exp(-L) \ne 0 $ . QED .

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Using the same "product-to-sum" strategy as @Did, we get
$${1\over 2}\prod_{k=2}^N\left({1-{1\over 2^k}}\right) \geq {1\over 2}\left(1-\sum_{k=2}^N{1\over 2^k}\right) ={1\over 4}+\left({1\over 2}\right)^{N+1}\geq {1\over 4}.$$

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+1. Simpler, clearer, more clever. – Did Nov 4 '11 at 9:15
@Didier Thanks for the kind words. – Byron Schmuland Nov 4 '11 at 12:13

We begin with a simple inequality. For every $x$ in $[0,\frac12]$, $\mathrm e^x\leqslant2$ hence $\frac12\leqslant\mathrm e^{-x}$. Integrating this on the interval $[0,x]$ yields $\frac12x\leqslant1-\mathrm e^{-x}$, that is, $1-\frac12x\geqslant\mathrm e^{-x}$.

Applying this to $x=\frac1{2^{k-1}}$ for each $k$ between $2$ and $n$, one gets $$ \prod_{k=1}^n\left(1-\frac1{2^k}\right)\geqslant\frac12\exp\left(-\sum_{k=2}^n\frac1{2^{k-1}}\right). $$ Since the sum in the exponential is less than $1$, all the partial products are $\geqslant\frac1{2\mathrm e}$ and $$ \prod_{k=1}^{+\infty}\left(1-\frac1{2^k}\right)\geqslant\frac1{2\mathrm e}>0. $$

Edit The convexity inequality $1-x\geqslant4^{-x}$, valid for every $x$ in $(0,\frac12]$, yields the lower bound $\frac14$.

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Of course, it is a general fact that, for any sequence $(a_k)_k\subset[0,1)$, the infinite product $\prod\limits_k(1-a_k)$ is positive if and only if the series $\sum\limits_ka_k$ converges. – Did Nov 3 '11 at 19:46

The following requires no tools other than induction. In particular, machinery from the calculus is not used.

Let $f(n)$ be the product up to the term $1-\dfrac{1}{2^n}$. We show by induction that $$f(n)\ge\frac{1}{4} +\frac{1}{2^{n+1}}.$$ The result is true at $n=1$. For the induction step, note that $$f(m+1)=f(m)\left(1-\frac{1}{2^{m+1}}\right)\ge\left(\frac{1}{4}+\frac{1}{2^{m+1}}\right)\left(1-\frac{1}{2^{m+1}}\right).$$ Expand the product on the right. We get $$\frac{1}{4}+\frac{1}{2^{m+1}}-\frac{1}{4}\cdot\frac{1}{2^{m+1}}-\frac{1}{2^{2m+2}}.\qquad(\ast)$$ Rewrite $\dfrac{1}{4}\cdot\dfrac{1}{2^{m+1}}$ as $\dfrac{1}{2^{m+2}}-\dfrac{1}{2^{m+3}}$. Then $(\ast)$ becomes $$\frac{1}{4}+\frac{1}{2^{m+2}}+\frac{1}{2^{m+3}}-\frac{1}{2^{2m+2}}.$$ The term $\dfrac{1}{2^{m+3}}-\dfrac{1}{2^{2m+2}}$ is non-negative. Thus $f(m+1)\ge\dfrac{1}{4}+\dfrac{1}{2^{m+2}}$, which completes the induction step.

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When you substitute $\frac{1}{2^{m+2}}-\frac{1}{2^{m+3}}$ in (*) it should become: $\frac{1}{4} – Chun-Yue Nov 4 '11 at 0:02
The $-(1/4)(1/2^{m+1})$ becomes $-1/2^{m+2}+1/2^{m+3}$. Then the $-1/2^{m+2}$ combines with the $1/2^{m+1}$ to the left of it to give $1/2^{m+2}$ (I skipped a step). – André Nicolas Nov 4 '11 at 0:17

I wonder if this works. It suffices to show that $\displaystyle \sum_{k=1}^{\infty}\log\left(1-\frac{1}{2^k}\right)>-\infty$. Note though that $\displaystyle \lim_{k\to\infty}\frac{\log\left(1-\frac{1}{2^k}\right)}{-\frac{1}{2^k}}=1$ and $\displaystyle \sum_{k=1}^{\infty}-\frac{1}{2^k}$ converges easily by the ratio test.

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I think you have that last formula wrong - as the terms go to $1$ as $k$ goes to infinity, the sum can't possibly converge. Did you mean just $-1\over 2^k$ instead of $1-{1\over 2^k}$? – Steven Stadnicki Nov 3 '11 at 20:11
Of course! Thank you Steven! – Alex Youcis Nov 3 '11 at 20:18

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