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

Note by Chinmay Sangawadekar 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.

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

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TopNewestSince \(\mathrm{gcd}(n,2015) \ge1\) 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.

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