Ytukyg

Let $S_n=prod_{k=3}^n (1-tan^4frac{pi}{2^k})$. What is the value of $lim_{n to infty} S_n$ ?

What I attempted:-

$log S_n=sum_{k=3}^n log (1-tan^4frac{pi}{2^k})$.

Since $lim_{x to 0} frac{tan x}{x}=1$, $tan^4frac{pi}{2^k}approx left(frac{pi}{2^k}right)^4$

Thus, $log S_n=sum_{k=3}^n log (1-frac{pi^4}{2^{4k}})approx sum_{k=3}^nleft( -frac{pi^4}{2^{4k}}right) $

Taking limit as $n to infty$, $lim_{n to infty} log S_n=frac{-pi^4}{3840}$.

Finally, $lim_{n to infty}S_n=e^{frac{-pi^4}{3840}}approx 1-frac{pi^4}{3840}$

Am I correct? Is there any other better method which yield the accurate limit? I was told that the exact limit should be one of the $4$ options:- $frac{pi^3}{4},frac{pi^3}{16}, frac{pi^3}{32},frac{pi^3}{256}$. I guess the third option is correct as it is giving almost the same value that I have obtained.

You obtained an approximation of the exact value. In order to find such exact vale, note that
$$(1-tan^4(alpha/2))=(1+tan^2(alpha/2))(1-tan^2(alpha/2))=
frac{4}{cos(alpha)}left(frac{tan(alpha/2)}{tan(alpha)}right)^2.$$
Hence, as $n$ goes to infinity,
$$prod_{k=3}^n left(1-tan^4(pi/2^k)right)=frac{4^{n-2}}{prod_{k=3}^ncos(pi/2^{k-1})}cdot left(prod_{k=3}^nfrac{tan(pi/2^k)}{tan(pi/2^{k-1})}right)^2\=4^{n-2}cdot 2^{n-2}sin(pi/2^{n-1})cdot left(frac{tan(pi/2^n)}{tan(pi/2^{2})}right)^2to frac{pi^3}{32}$$
where we used the known fact that
$$prodlimits_{k=2}^{n}cosleft(frac{pi }{2^{k}}right)=
frac{1}{2^{n-1}sin(pi/2^n)}$$
(see for example How to evaluate $limlimits_{nto infty}prodlimits_{r=2}^{n}cosleft(frac{pi}{2^{r}}right)$).

What you're describing is how to arrive at an approximation of the actual product.
But you can compute the exact value for the product using trigonometric identities.

$$begin{split}
S_n&=prod_{k=3}^n (1-tan^4frac{pi}{2^k})\
&=prod_{k=3}^n (1-tan^2frac{pi}{2^k})(1+tan^2frac{pi}{2^k})\
&=prod_{k=3}^n (frac {cos^2(frac{pi}{2^k}) - sin^2(frac{pi}{2^k})}{cos^2 frac{pi}{2^k}})(frac 1 {cos^2frac{pi}{2^k}})\
&=prod_{k=3}^n frac {cos(frac {pi}{2^{k-1}})}{cos^4 frac{pi}{2^k}}\
&= frac{cosfrac pi 4}{cos^4 frac pi 8}frac{cosfrac pi 8}{cos^4 frac pi {16}}frac{cosfrac pi 16}{cos^4 frac pi {32}}...\
&=cosfrac pi 4 prod_{k=3}^n frac 1 {cos^3 frac pi {2^k}}
end{split}$$
Now because $$frac 1 {cos theta}=frac {2 sin theta}{sin 2theta}$$
you can verify that, for any $0<theta<pi$,
$$
prod_{k=1}^{+infty} frac 1 {cos frac theta {2^k}}=frac theta {sin theta}
$$
Which yields, for $theta=frac pi 4$,
$$
prod_{k=3}^{+infty} frac 1 {cos frac pi {2^k}}=frac pi {4sin frac pi 4}
$$
So finally,
$$lim_{nrightarrow+infty}S_n=left(cos{frac pi 4}right)frac {pi^3} {4^3sin^3 frac pi 4}$$
In other words,
$$lim_{nrightarrow+infty}S_n=frac {pi^3}{32}$$