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show that $$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge\dfrac{14}{25}\tag{1}$$

My try: I only prove following not strong inequality:

$$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge\dfrac{1}{2}$$

proof: use Bernoulli inequality $$(1+x_{1})(1+x_{2})\cdots (1+x_{n})\ge 1+x_{1}+x_{2}+\cdots +x_{n},$$where $x_{i}\ge -1$

so

$$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge1-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{3^n}\right)$$ so $$LHS\ge1-\sum_{n=1}^{\infty}\dfrac{1}{3^n}=1-\dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}=\dfrac{1}{2}$$

But for $(1)$,I can't prove it,Thank you

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    $\begingroup$ But $\left(1 - \frac13\right)$ is already less than $\frac{24}{25}$, and the products decrease as $n$ becomes larger, so the thing you're trying to prove is obviously false. $\endgroup$ – MJD Oct 30 '13 at 3:36
  • $\begingroup$ Seriously @MJD, they're getting larger. $\endgroup$ – user28877 Oct 30 '13 at 3:38
  • $\begingroup$ Yes, @MJD, I agree. I read that as the terms getting larger. But yes the partial products are getting smaller. $\endgroup$ – user28877 Oct 30 '13 at 3:41
  • $\begingroup$ Incidentally, computer calculation shows that the product converges to approximately 0.560126077927948. $\endgroup$ – MJD Oct 30 '13 at 3:42
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    $\begingroup$ sorry,I have edit,Thank you everyone. $\endgroup$ – china math Oct 30 '13 at 3:45
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Even better we can compute an excellent approximation by the standard repertoire of functions of a complex variable to the value of this product, call it $P$. This has a certain didactic value. Introduce $$S = \log P =\sum_{n\ge 1} \log\left(1-\frac{1}{3^n}\right).$$ The value $S$ is taken on at $x=1$ of the following harmonic sum: $$S(x) = \sum_{n\ge 1} \log\left(1-\frac{1}{3^{xn}}\right).$$ Now we will evaluate this by inverting its Mellin transform.

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = 1, \quad \mu_k = k \quad \text{and} \quad g(x) = \log\left(1-\frac{1}{3^x}\right).$$ We need the Mellin transform $g^*(s)$ of $g(x)$ which is $$\int_0^\infty \log\left(1-\frac{1}{3^x}\right) x^{s-1} dx = \left[\log\left(1-\frac{1}{3^x}\right) \times \frac{x^s}{s} \right]_0^\infty - \int_0^\infty \frac{\log 3}{3^x-1} \frac{x^s}{s} dx .$$ The first term in the difference vanishes for $\Re(s)\ge 1.$ It remains to do the calculation of the integral, which goes like this: $$\int_0^\infty \frac{1}{3^x-1} x^s dx = \int_0^\infty \frac{3^{-x}}{1-3^{-x}} x^s dx = \int_0^\infty 3^{-x}\sum_{q\ge 0} 3^{-qx} x^s dx = \sum_{q\ge 0} \int_0^\infty 3^{-(q+1)x} x^s dx.$$ Continuing, we find $$\sum_{q\ge 0} \int_0^\infty e^{-(q+1)\log 3 \times x} x^s dx = \sum_{q\ge 0} \frac{1}{(\log 3\times (q+1))^{s+1}}\Gamma(s+1) = \frac{\Gamma(s+1)\zeta(s+1)}{(\log 3)^{s+1}}.$$ This gives that $$g^*(s) = - \frac{\log 3}{s} \frac{\Gamma(s+1)\zeta(s+1)}{(\log 3)^{s+1}} = - \frac{1}{s} \frac{\Gamma(s+1)\zeta(s+1)}{(\log 3)^s}.$$ By the harmonic sum identity we thus have for $Q(s)$, the Mellin transform of $S(x)$, that $$Q(s) = - \frac{1}{s (\log 3)^s} \Gamma(s+1)\zeta(s)\zeta(s+1).$$ Now inverting the transform in the left half-plane to the left of $\Re(s)=3/2$ for an expansion about zero we find that the trivial zeros of the two zeta function terms cancel the poles of the gamma function, so that we are left with just three poles. We get $$\begin{align} \mathrm{Res}(Q(s)/x^s; s=1) &= -\frac{1}{6} \frac{\pi^2}{\log 3} \frac{1}{x},\\ \mathrm{Res}(Q(s)/x^s; s=0) &= \frac{1}{2} \log \frac{2\pi}{\log 3} - \frac{1}{2}\log x, \quad\text{and}\\ \mathrm{Res}(Q(s)/x^s; s=-1) &= \frac{1}{24} x \log 3. \end{align}$$ Setting $x=1$ we obtain an approximation of $S(1)$ which is good to fifteen digits, namely $$S(1) = S \approx -\frac{1}{6} \frac{\pi^2}{\log 3} + \frac{1}{2} \log \frac{2\pi}{\log 3} + \frac{1}{24} \log 3\\ \approx -0.57959338143590694$$ whereas the exact value is $-0.57959338143590718.$ This finally gives the following approximate formula for $P$ (also good to fifteen digits): $$ P \approx \exp\left(-\frac{1}{6} \frac{\pi^2}{\log 3}\right) \times \sqrt{\frac{2\pi}{\log 3} } \times 3^{1/24} \approx 0.56012607792794900$$ whereas the exact value is $0.56012607792794894.$

It seems interesting to ask whether there is a better approximation using the standard repertoire than the one given above and it might be an instructive exercise (and produce a better approximation) to extract the first few terms of the sum $S(x)$ (using their exact values) and invert the Mellin transform of the rest.

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  • $\begingroup$ Beautiful demonstration ! $\endgroup$ – Claude Leibovici Oct 31 '13 at 7:22
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You can use the same idea, but first pull out a couple of factors from the product: $$(2/3) (8/9) (26/27) \left(1-\sum_{j=4}^\infty 1/3^j\right)=11024/19683>14/25.$$

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  • $\begingroup$ That $=$ sign should be $\ge$, shouldn't it? $\endgroup$ – MJD Oct 30 '13 at 4:16
  • $\begingroup$ @MJD No, but I do not begin with the OP's original expression but rather a lower bound. $\endgroup$ – user940 Oct 30 '13 at 11:49
  • $\begingroup$ Right, thanks! ${}{}$ $\endgroup$ – MJD Oct 30 '13 at 12:40
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Here's another, somewhat complicated way to prove things. We can rewrite the inequality (with $n$ taken to infinity) as

$${25\over14}\ge\prod_{k=1}^\infty{1\over1-(1/3)^k}=\sum_{k=0}^\infty p(k)(1/3)^k$$

where $p(k)$ is the partition function $1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56,77,101,\ldots$. Let's accept for the moment the inequality

$$p(k)\le{1\over50}2^k\text{ for } k\gt11$$

Then we have

$$\sum_{k=0}^\infty p(k)(1/3)^k \le 1+{1\over3}+{2\over3^2}+{3\over3^3}+{5\over3^4}+{7\over3^5}+{11\over3^6}+{15\over3^7}+{22\over3^8}+{30\over3^9}+{42\over3^{10}}+{56\over3^{11}}+{3\over50}\left({2\over3}\right)^{12}$$

where the final term comes from

$$p(12)(1/3)^{12}+\cdots \le {1\over50}\left({2\over3}\right)^{12}\left(1+{2\over3}+\left({2\over3}\right)^2+\cdots\right)={1\over50}\left({2\over3}\right)^{12}{1\over1-(2/3)}={3\over50}\left({2\over3}\right)^{12}$$

The sum of all those fractions is not a lot of fun to work out, but it turns out to give $$\sum_{k=0}^\infty p(k)(1/3)^k \le {316217\over177147}+{3\over50}\left({2\over3}\right)^{12}\approx1.7855\lt1.7857\approx{25\over14}$$

The inequality on partition numbers that made this work can be proven by induction from Euler's pentagonal number theorem:

$$\begin{align} p(k)&=p(k-1)+p(k-2)-p(k-5)-p(k-7)+p(k-12)+p(k-15)-\cdots\cr &=p(k-1)+p(k-2)-\left(p(k-5)-p(k-12)\right)-\left(p(k-7)-p(k-15)\right)-\cdots\cr &\le p(k-1)+p(k-2)\cr \end{align}$$

Clearly one can conclude from this an inequality of the form $p(k)\le cr^k$ with any ratio $r$ greater than or equal to the golden ratio $\phi=(1+\sqrt5)/2$, but with a constant $c$ and/or starting value for $k$ that need to be checked against the initial terms of the sequence. To the extent that I played around with different possibilities, I always found I had to go out to the twelfth term to make things less than $25/14$.

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The product corresponds to QPochhammer[1/3,1/3,m] which is a decreasing function. For m going to infinity, its limit is QPochhammer[1/3,1/3] which is approximately 0.5601260779279490 as mentioned by MJD (14/25 = 0.56000).

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