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Is it possible to evaluate?


Note by Chinmay Sangawadekar
1 year, 3 months ago

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Since \(\mathrm{gcd}(n,2015) \ge1\) and \(\mathrm{gcd}(n,2016)\le2016\), we have \[ \frac{\mathrm{gcd}(n,2015)}{\mathrm{gcd}(n,2016)} \; \ge \; \tfrac{1}{2016} \] for all \(n\), and so the infinite series must diverge. Mark Hennings · 1 year, 3 months ago

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