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Maybe most of you have seen this before, but I find it cool and that's why I thought I share it with you.

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

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)

[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)

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)

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