Question

Can anyone show me the sum of this series:

Answer 1

\[\sum_{n=0}^{\infty}\frac{ p _{n} }{ 2^{n} }\]

Answer 2

p is the nth prime

Answer 3

there are a bounch of the prime number theorems that states there is a constant \(c\) such that \[p_{n}\leq cn\log n \]for all \(n\ge 2\). So your sum is bounded from above by\[\sum_{n=0}^{\infty}\frac{ p _{n} }{ 2^{n} }\le \frac{7}{2}+c\sum_{n=2}^\infty\frac{n\log n}{2^{n}}\leq\frac{7}{2}+c\sum_{n=2}^\infty\frac{n^{2}}{2^{n}}=\frac{7}{2}+\frac{11}{2}c\]As to the value, it seems unlikely that it is expressible in terms of common constants.

Answer 4

I'm seeing a lot of "math processing error" in red. Is this permanent?

Answer 5

that happens sometimes...how about now :)