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