How can they compute infinite products?

Through the use of logic and analysis.

Let's take a very simple case: \prod_{n = 1}^\infty 1^n. As you probably already know, 1^x = 1 always. And you already know that 1 times 1 is 1. So what do you get you multiply an infinity of 1s? Just 1. This is an infinite product you can compute quite straightforwardly.

Now let's look at a more interesting example: \prod_{n = 2}^\infty (1 – 1/n^2). Our first multiplicand is 0.75 (1 minus a quarter is 3/4). We multiply that by 1 – 1/9, then by 1 – 1/16, and then by 1 – 1/25, etc. The first few values are 0.6666..., 0.625, 0.6, 0.583333..., 0.571428..., and so on and so forth. Each new multiplicand is getting closer and closer to 1. So, the running product is gradually settling down. If you take this far enough on your calculator, you should get a result fairly close to 0.5.

Added again: Phi does NOT give the largest possible value. Phi+.01 seems to be larger.

Added: I guess I cannot quite say that this is the largest value, but it looks like it is the value when phi is the golden ratio (or at least very close). It looks suspiciously like the maximum.

phi has the property that every power and all of the fractions in this product can be written using Fibonacci numbers. However, this has not helped me to simplify the result.

You miss-stated the question!

Yes, it is the value for phi, as stated above. See A062073 at link below for more.

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The value 1.226742 approximates the LARGEST value that can be attained with this infinite product and phi > 1, which is needed for convergence.

I went to the link below and entered "plot y=(1+1/x^2)(1-1/x^4)(1+1/x^6)(1-1/x^8) for x between 1 and 4" and got a beautiful graph!

"maximize y=(1+1/x^2)(1-1/x^4)(1+1/x^6)(1-1/x^8) for x=1 to 4" gave answer at x=1.608, so maybe phi=golden ratio is the answer for the infinite version!

There is no way to compute infinite products. That's why they're called infinite.