Sum of a Sequence based on prime numbers I was playing around with prime numbers and arrived at this sequence:
$$ 
\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^{11}}+\frac{1}{2^{13}}+\frac{1}{2^{17}}+\frac{1}{2^{19}}+\frac{1}{2^{23}}+\frac{1}{2^{27}} ... = 0.414682388306 \approx \sqrt{2}-1
$$ 
Although the sum does not approach $\sqrt2 - 1$ it is very similar. Is there any reason or is this just a coincidence.
 A: Here is an explanation why $\dfrac{209}{504}$ is a much better approximation than $\sqrt{2} - 1$ and also shows why the latter must be a co-incidence.
The number $\sum_{n = 1}^{\infty} \frac{1}{2^{p_n}}$ is known as the prime constant and its value is approximately $0.414682509...$.
$$
s = \sum_{n = 1}^{\infty} \frac{1}{2^{p_n}} < \frac{1}{2^2} + \frac{1}{2^3}
+ \sum_{n = 1}^{\infty} \frac{1}{2^{6n-1}} + \sum_{n = 1}^{\infty} \frac{1}{2^{6n+1}} = \frac{3}{8} + \frac{5}{126} = \frac{209}{504} \approx 0.41468253...
$$
Each term in the RHS decreases roughly by a factor of $1/64$ and among the first few natural numbers, we have a low density of numbers of the form $6n \pm 1$ which are not primes; $25$ is the only number of this form below $30$ which is not a prime. Hence the actual sum must be very close to the above upper bound because $1/64$ raised to non primes of the $6n \pm 1$ will not effect the first few decimal places.
In fact $\dfrac{209}{504}$ agrees to seven decimal places with the actual sum and is therefore a much better estimate that $\sqrt{2}-1$ which agrees only to $3$ decimal places.
