The “many more days that include them” is the 3^n part in my expression that is missing from any per day series. This 3^n is the sum of all interviews in that coin flip sequence (“coin flip sequence” = “all the interviews that are done because one coin flip showed up tails”, right?) and in the per day (aka per interview) series the exact same sum exists, just as 3^n summands.
In both cases, the weight of the later coin flip sequences increases, because the number of interviews (3^n) increases faster than the probabilistic weight of the coin flip (1/2^n) decreases.
However, this doesn’t mean that there exists no Cesàro sum. In fact the existence of such a sum can be proven for my original expression because the quotient of the last two numerators (if we include both odd and even coin flips) of the isomorphic series is always 3:1, regardless of wether the last coin flip was even or odd. (The same thing can be said for the quotient of the last 3^n and 3^(n-1) summands of your series. Basically, the per day series is just a dragged out per coin flip series.)
The reason why my estimation for the Cesàro sum is 0.5 is that if we express that quotient in a way that one coin state is written first, then it alternates between 3:1 and 1:3, which results in 1:1 which is 0.5. Obviously this is not exact maths, but it’s a good way for a quick estimation. (Alternatively, you could intuitively infer that if there exists a Cesàro sum it must be 0.5, because whether you look for even or odd coin flips gets increasingly irrelevant as the series approaches infinity.)
Also, since I haven’t previously touched upon the subject of the isomorphic series: If we call my original expression f, then we can construct a function g where g(n) = f(n)-f(n-1) with f(-1) = 0, and a series a = g(0) + g(1) + g(2) + …
The “many more days that include them” is the 3^n part in my expression that is missing from any per day series. This 3^n is the sum of all interviews in that coin flip sequence (“coin flip sequence” = “all the interviews that are done because one coin flip showed up tails”, right?) and in the per day (aka per interview) series the exact same sum exists, just as 3^n summands.
In both cases, the weight of the later coin flip sequences increases, because the number of interviews (3^n) increases faster than the probabilistic weight of the coin flip (1/2^n) decreases.
However, this doesn’t mean that there exists no Cesàro sum. In fact the existence of such a sum can be proven for my original expression because the quotient of the last two numerators (if we include both odd and even coin flips) of the isomorphic series is always 3:1, regardless of wether the last coin flip was even or odd. (The same thing can be said for the quotient of the last 3^n and 3^(n-1) summands of your series. Basically, the per day series is just a dragged out per coin flip series.)
The reason why my estimation for the Cesàro sum is 0.5 is that if we express that quotient in a way that one coin state is written first, then it alternates between 3:1 and 1:3, which results in 1:1 which is 0.5. Obviously this is not exact maths, but it’s a good way for a quick estimation. (Alternatively, you could intuitively infer that if there exists a Cesàro sum it must be 0.5, because whether you look for even or odd coin flips gets increasingly irrelevant as the series approaches infinity.)
Also, since I haven’t previously touched upon the subject of the isomorphic series: If we call my original expression f, then we can construct a function g where g(n) = f(n)-f(n-1) with f(-1) = 0, and a series a = g(0) + g(1) + g(2) + …
Does that all make sense?