# Is it possible to evaluate?

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

2 years, 1 month 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.

- 2 years, 1 month ago

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