A very difficult task that...

[AlfredFlaßhaar]

...not to be forgotten. It comes from the Irish MO of 1990, Paper 2, Problem 9, and, as a student-friendly problem, is one of the most difficult I've ever encountered:
Let t be a real number, and let (a with index n) a_n = 2*cos(t/(2^n)), n=1, 2, ...
Let b_n be the product a_1*a_2*...*a_n. Simplify the term b_n so that it does not contain a product of n factors, and prove lim(n-->00) b_n = (2*cos(t)+1)/3 .

[Werner_E]

And the question regarding Mathcad here is ... which one?
Whether Mathcad's symbolics is capable of simplifying the term b_n(t) accordingly and specifying the limit value?
I would guess - with a high probability “no”.

By the way, there seems to be something wrong with your statement, as you can easily see if you look at t=0.
For t=0, b_n=2ⁿ (i.e. grows for n-->oo over all limits). However, when inserted into the limit function you specified, t=0 gives the value 1.

The term you name for b_n could be simplified to b_n=sin(t)/sin(t/2ⁿ) by dividing b_n by sin(t/2ⁿ) and continuously applying the theorem of addition -> 2*cos(t/2^k)*sin(t/2^k) = sin(t/2^k-1).

But also this simplified b_n=sin(t)/sin(t/2ⁿ) goes beyond all limits if n->oo, doesn't it?

I can't see a limit of (2*cos(t)+1)/3 here.

Maybe it helps if you post a picture of the original task or provide a link.

[Werner_E]

After playing around with the problem a bit, I now think that you forgot to enter a summand -1 when you provided a_n, right?
That would make the limit function correct and, as I suspected, Mathcad's symbolic does not succeed in either simplifying the product nor in calculating its limit.

But at least we can 'show' by plotting and visual inspection that we seem to have a rather fast convergence:

And yes, I also think that it is indeed quite a difficult task.

[AlfredFlaßhaar]

Yes, the "-1" was forgotten - sorry.