We know that $$ \log{2} = \sum_{n=1}^{\infty} {\frac{1}{2n (2n-1)}} \space < \space \sum_{p \space prime} {\frac{1}{p (p-1)}} \space < \space \sum_{n=2}^{\infty} {\frac{1}{n (n-1)}} = 1 $$ Is there any formula for the mid sum? (constant, value, limit, integral, function)
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3Why is the first inequality true? – Wojowu Sep 08 '16 at 15:00
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I have added the (closed-form) tag, since it seems that this is what you are asking for. – Wojowu Sep 08 '16 at 15:01
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(1) $ \log{2} < (\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{4.5}=\frac{43}{60}=0.717) $. (2) Any idea about the progression, $ 2n(2n-1)={1.2,3.4,5.6,...} \space,,\space p(p-1)={1.2,2.3,4.5,...} \space,,\space n(n-1)={1.2,2.3,3.4,...} $. – Hazem Orabi Sep 10 '16 at 21:00