Notice that your numbers appear in denominators only, so let r_i = 1/n_i.

Then you've written r_1 + r_2(1 - 2r_1) + r_3(1 - 2r_1)(1 - 2r_2) + ...

presumably with r_i in (0, 1/2).

Let f(r_1, r_2, ...) = r_1 + r_2(1 - 2r_1) + r_3(1 - 2r_1)(1 - 2r_2) + ...

Thus, f(r_1, r_2, ...) = r_1 + (1-2r_1)f(r_2, r_3, ...)

Claim: for any r in (0, 1/2), we want r + (1-2r)x to be closer to 2 than x is.

Is this likely? If x > 2, we'd like this to be smaller than x:

r + (1-2r)x < x iff r < (2r)x iff 1/2 < x.

Thus it's decreasing for any x > 1/2.

So the next thing we'd like to do is show that it's greater than 2. Suppose x > 2. Must it be that r + (1-2r)x > 2?

r + (1-2r)x > 2 iff r + x - 2rx > 2 iff r(1 - 2x) > 2 - x.

We know that 1-2x < 0, so the above is equivalent to r < (2-x)/(1-2x). So the new claim is that this value is a least 1/2.

Claim: (2-x)/(1-2x) < 1/2. Well, this is equivalent to 4-2x > 1-2x, since 1-2x is negative. And this is because equivalent to 3>0, which is true.

Thus your claim is actually true:

If I take some x0 > 2, and compute x{n+1} = r + (1-2r)x_n, the limit will be 2.