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I am Anay, here is a problem I am stuck with:

$$x = \prod\limits_{n=1}^{\infty }\left ( 1 + \frac{1}{3^n} \right )$$

The task is to find the value of $x$. (obviously, we aren't supposed to have infinite products or sums, etc. in the answer)

This is what I have done:

We define the sequence $a_{k}$ as,

$$a_{k} = \prod\limits_{n=1}^{k }\left ( 1 + \frac{1}{3^n} \right )$$

First, we put some bounds on $a_{k}$, as it is a increasing sequence, we already have the lower bound as $\frac{4}{3}$. Now to get the higher bound, we have the following inequality for all integers $x$ (easily proved through binomial expansion):

$$\left(1 + \frac{1}{x}\right)^{x} < e$$

So,

$$(1+x) < e^{\frac{1}{x}}$$

Using this inequality many times, we have, (using the formula for sum of a geometric progression)

$$a_{k} < e^2$$

Thus,

$$\frac{4}{3}\leq a_{k}< e^2$$

Then, to prove that this sequence is converging we show that it has the Cauchy Property. This can be done as follows:

First, we have,

$$a_{k} = a_{k-1} + \frac{a_{k-1}}{3^k}$$

So adding such equations for $a_{1}$, $a_{2}$ ..... $a_{k}$, we see that all terms cancel out and the following remains:

$$a_{k} = a_{1} + \sum\limits_{i = 1}^{k-1}\frac{a_{i}}{3^{i+1}}$$

So, if $m < n$,

$$a_{n} - a_{m} = \sum\limits_{i=m}^{n-1} \frac{a_{i}}{3^{i+1}} < a_{n-1}\left(\frac{3^{n} - 3^{m}}{3^{m+n}\times2 }\right)$$

As $a_{k}$ is a increasing sequence, we have used $a_{n-1} > a_{n-2}>....>a_{m} $. And then we use the formula for sum of a geometric progression to get the result. Now we can see that when $m$ and $n$ are large enough, we can have the RHS arbitrarily small as $a_{n-1}$ has a upper bound ($e^2$), thus the sequence has the Cauchy property and it is converging.

After this I thought may be the sequence converges to the bound which I established ($e^2$), but it is not so as I checked it through a computer program, it approaches around $1.56$, which is far below $e^2$. So, after this I try many other methods to find where the sequence converges to but I found no luck. Also, I couldn't find any results on Google, so I have come to your help. How do I solve this?

Thanks in advance.

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    $\begingroup$ Wolfram Alpha thinks the result is ≈1.564934018567011537938849106728835416569425919895035009496... $\endgroup$ – MatheMagic Dec 27 '16 at 18:06
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    $\begingroup$ See: en.wikipedia.org/wiki/Euler_function and en.wikipedia.org/wiki/Q-Pochhammer_symbol In summary: there are no simple closed form for this product in elementary functions. Try to approximate it instead. $\endgroup$ – Winther Dec 27 '16 at 18:13
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    $\begingroup$ $\displaystyle\prod_{n=1}^\infty\left(1+\frac{1}{3^{n}}\right)=\frac{\left(-1;\frac{1}{3}\right)_{\infty}}{2}$. $\endgroup$ – MatheMagic Dec 27 '16 at 18:20
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    $\begingroup$ See the link:math.stackexchange.com/questions/1924882/… $\endgroup$ – MatheMagic Dec 27 '16 at 18:24
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    $\begingroup$ A good approximation is $$\prod_{n=1}^\infty\left(1+\frac{1}{3^n}\right) \approx e^{\frac{1}{2\cdot 3^k}}\prod_{n=1}^{k}\left(1+\frac{1}{3^n}\right)$$ for any integer $k$. The higher $k$ the better the approximation. For example for $k=4$ it gives $\frac{91840 \sqrt[162]{e}}{59049} \simeq 1.56495$ while the exact answer is $\simeq 1.56493$. $\endgroup$ – Winther Dec 27 '16 at 18:25
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$$\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{3^2}\right)=1+\frac13+\frac1{3^2}+\frac1{3^3}$$

$$\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{3^2}\right)\left(1+\dfrac{1}{3^3}\right)=1+\frac13+\frac1{3^2}+\frac2{3^3}+\frac1{3^4}+\frac1{3^5}+\frac1{3^6}$$

I guess continuing this you can observe a pattern and it will leads to the solution.
Good luck

EDIT:
Above pattern shows us, the infinite summation can be written as $$\sum_{n=1}^{\infty}\dfrac{q(n)}{3^n},$$ where $q$ is the partition function. So there is no closed form involving elementary functions.

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  • $\begingroup$ There is no simple pattern that allows a simple solution here. The solution can formally be written in terms of en.wikipedia.org/wiki/Q-Pochhammer_symbol or en.wikipedia.org/wiki/Euler_function - these are non-elementary functions (defined as infinite products) and do not have a simple representation. $\endgroup$ – Winther Dec 27 '16 at 18:28
  • $\begingroup$ Yes. Just now I realize that the pattern involves the partition function and it is non-elementary. $\endgroup$ – Bumblebee Dec 27 '16 at 18:31
  • $\begingroup$ uh? I don't see how this pattern can help in a infinite sequence $\endgroup$ – Anay Karnik Dec 27 '16 at 18:32
  • $\begingroup$ We define Infinite products as the limit of a finite product. So there is a relationship between this pattern and your infinite product. :) $\endgroup$ – Bumblebee Dec 27 '16 at 18:35

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