Not a Coincidence

@Akhmad Syaikhu Firizal posted a great solution using Arithmetic-Geometric Progression , I'll use the famous Euler's Prime Generating Polynomial

Euler showed that the polynomial $n^2 -n +41$ gives prime numbers for integers $-40\leq n\leq 40$ .

Now realize that the fraction is just $\large{\sum_{n=1}^{\infty} \dfrac{n^2 -n +41}{10^{4n}}=\dfrac{4099200041}{999700029999}}$

Since the polynomial will produce prime numbers till $n=40$ only. The first number not prime is $\boxed{n=41}$

We may as well want to check that the $n^2 -n +41$ produces a prime of 4 digits only since a larger digited prime may result in the carry. But, $(40)^2 - 40+41 = 1523$ which is 4-digit prime.

$0.0041004300470053....=0.004100410041....+2(0.0000000100020003....+0.00000000000100020003....+.....)$ where, $0.004100410041...= \frac{41}{10^4-1}$ then, $0.000100020003....= \frac{1}{10^4} + \frac{2}{10^8} + \frac{3}{10^12} +.... = \frac{10^4}{(10^4-1)^2}$ so, $0.0041004300470053....= 0.004100410041.... + 2(0.0000000100020003.... + 0.00000000000100020003.... + .....)$ $= \frac{41}{10^4-1} + 2( \frac{10^4}{(10^4)(10^4-1)^2} + \frac{10^4}{(10^8)(10^4-1)^2} + \frac{10^4}{(10^12)(10^4-1)^2} + ...)$ $= \frac{41}{10^4-1} +2(\frac{10^4}{(10^4-1)^3}) = \frac{41(10^4-1)^2+2(10^4)}{(10^4-1)^3} = \frac{4099200041}{999700029999}$ so it is clear that this is not a coincidence, and if we eliminate zeros , we will get a sequence of numbers that is $41,43,47,53,....$ where, $43= 41+2$ $47=41+2+4$ $53=41+2+4+6$ and so forth. $a_{n} = 41+2+4+6+....+2(n-1) = 41+2(1+2+3+....+(n-1)) = 41+2\frac{(n-1)(n)}{2} = 41+(n-1)(n)$ the function of pattern is $n(n-1)+41$ so, this function will not generate a prime number , if n ( n - 1 ) can be divided by 41 , so n that meet is 41 and 42.the least n is 41. $$ then, this pattern will cease after 40 prime number. $$ so the answer is 40