Begin with the Von Mangoldt divisor identitySumming it for and reversing the order givesThe given bound and the Stirling formula therefore imply
Apply partial summation with the weight . Writing , where , givesIndeed, the integral of converges, and its tail from to infinity is .
Grouping the left side by prime powers yieldsThe full double series over converges. Its tail beyond is : split at , use a geometric series for , and compare with the corresponding sum over integers. Absorbing its limit into the constant proves the Mertens theorem for reciprocal primes
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