Maybe most of you have seen this before, but I find it cool and that's why I thought I share it with you.(adsbygoogle = window.adsbygoogle || []).push({});

Lets look at the sum

[tex]\sum_{p\leq N}\frac{1}{p}[/tex]

where p represents only prime numbers. If I calculate the value for this sum when taking into account all prime numbers < 10^{5}the result is 2.705 and taking into account all prime numbers < 10^{7}the result still is only 3.041. The obvious question now is: does this series converge to a fixed value, or does it diverge? Surprisingly (for me the first time I saw it), this series diverges for [tex]N\rightarrow\infty[/tex]!

The proof was already known by Euler and goes like this:

Consider

[tex]\prod_{p\leq N}\left(1-\frac{1}{p}\right)^{-1}[/tex]

where the product goes over all prime numbers [tex]p\leq N[/tex]. Now we can use the geometrical series for (1 - 1/p)^{-1}= 1 + 1/p + 1/p^{2}+ ... to find that

[tex]\prod_{p\leq N}\left(1-\frac{1}{p}\right)^{-1}=[/tex][tex]\prod_{p\leq N}\left(1+\frac{1}{p}+\frac{1}{p^2}+... \right)[/tex]

(continued in post II)

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Sum of 1/primes for all primes

**Physics Forums | Science Articles, Homework Help, Discussion**