I want to ask if there is this result $$\sum_{p\text{ are prime numbers}}\dfrac{1}{p^2}\approx 0.452224742...<\dfrac{1}{2}?$$ I sincerely thank you.
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Related. – Jan 02 '20 at 02:51
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2@Gae.S. Note the related MO post you provided is also linked to in the first question comment of my proposed duplicate Math SE question. – John Omielan Jan 02 '20 at 02:56
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3Note I found a likely better duplicate to use of How to prove $\sum_p p^{-2} < \frac{1}{2}$?. – John Omielan Jan 02 '20 at 03:00
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Wow, great! I sincerely thank you very much. Good health! – Tran Nam Son Jan 02 '20 at 03:33
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1Use that $$\sum_{p \ge k} p^{-2}\le\int_{k-1}^\infty x^{-2}dx= (k-1)^{-1}$$ taking $k=17$ gives your result. – reuns Jan 02 '20 at 07:33