how to solve $\lim_{n\to\infty} \prod_{i=1}^n (1+\frac{1}{3^{2i-1}})$?

Question

how to solve $\lim_{n\to\infty} \prod_{i=1}^n (1+\frac{1}{3^{2i-1}})$?

i’ve tried this:

let

$$ a_n= \prod_{i=1}^n (1+\frac{1}{3^{2i-1}}) $$

then

$$ lna_n=\sum_{i=1}^nln(1+ \frac{1}{3^{2i-1}})<\sum_{i=1}^n \frac{1}{3^{2i-1}}=\frac{1-\frac{1}{3^{2n}}}{2} $$

which means

$$ \lim_{n\to\infty}lna_n<\lim_{n\to\infty} \frac{1-\frac{1}{3^{2n}}}{2}=3/8 $$

$$ \lim_{n\to\infty} a_n<e^{3/8} $$

and

$$ lna_n=\sum_{i=1}^nln(1+ \frac{1}{3^{2i-1}})>\sum_{i=1}^n \frac{1}{1+3^{2i-1}}=? $$

but i don’t know how to sum up the right part of the inequation .

furthermore, even we got the number(wolframalpha told me is approximately 0.29) , the answer is 3/2, which means my solution is wrong at first, my solution only could prove the limit exists.

how to solve this problem?

Answer 1

Using the q-Pochhammer symbol the infinite product is $(-\frac13,\frac19)_{\infty}\approx 1.389120.$ It does not have a simpler closed form and is probably a transcendental number.