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Is it possible to evaluate?

\[\sum_{n=1}^{\infty}\frac{\gcd(n,2015)}{\gcd(n,2016)}\]

Note by Chinmay Sangawadekar
3 months, 2 weeks 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 · 3 months, 2 weeks ago

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