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):

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


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

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

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


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

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.

  • 1
    $\begingroup$ Wolfram Alpha thinks the result is ≈1.564934018567011537938849106728835416569425919895035009496... $\endgroup$ – MatheMagic Dec 27 '16 at 18:06
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ $\displaystyle\prod_{n=1}^\infty\left(1+\frac{1}{3^{n}}\right)=\frac{\left(-1;\frac{1}{3}\right)_{\infty}}{2}$. $\endgroup$ – MatheMagic Dec 27 '16 at 18:20
  • 5
    $\begingroup$ See the link:math.stackexchange.com/questions/1924882/… $\endgroup$ – MatheMagic Dec 27 '16 at 18:24
  • 1
    $\begingroup$ 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



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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.